integration involving fraction

[fresh_42]

Where did you get $x+1$ in the denominator from? You correctly split the quotient
$$\dfrac{1}{x(x^2+1)} =\dfrac{1}{x}-\dfrac{x}{x^2+1}$$This can be solved by logarithms, the first one is directly a logarithm, and the second one needs a substitution $u=x^2+1.$

Did you switch to your original problem at this stage without mention?

The original problem has a nasty anti-derivative with logs and arcus tangent.

$$\int \dfrac{x-2}{x^2-x+1}\, dx = \dfrac{1}{2} (\log(x^2 - x + 1) - 2 \sqrt{3} \tan^{-1}((2 x - 1)\sqrt{3})) + constant$$
You could probably do it by looking up the anti-derivatives of $\dfrac{1}{x^2-x+1}$ and $\dfrac{x}{x^2-x+1}$ eventually after another substitution. However, it is a definite integral. It is easier to solve by the residue method than by partial fraction decomposition. Did you read the article I linked? It contains an example of how it is done, although it only mentions that $\int (x^3+1)^{-1} dx$ can be solved accordingly.

Allow me a remark: stop using $\infty$ as if it were a number, it is not! $\int_0^\infty \dfrac{1}{x+1}\,dx$ does not converge! This is another problem with the partial fraction decomposition: it leaves you with two divergent integrals! Since $\int_0^\infty (x^3+1)^{-1} dx$ is finite, the two "divergencies" cancel each other what cannot be seen by the partial fractions.

 

[logistic_guy]

Where did you get $x+1$ in the denominator from? You correctly split the quotient
$$\dfrac{1}{x(x^2+1)} =\dfrac{1}{x}-\dfrac{x}{x^2+1}$$This can be solved by logarithms, the first one is directly a logarithm, and the second one needs a substitution $u=x^2+1.$

Did you switch to your original problem at this stage without mention?

The original problem has a nasty anti-derivative with logs and arcus tangent.

$$\int \dfrac{x-2}{x^2-x+1}\, dx = \dfrac{1}{2} (\log(x^2 - x + 1) - 2 \sqrt{3} \tan^{-1}((2 x - 1)\sqrt{3})) + constant$$
You could probably do it by looking up the anti-derivatives of $\dfrac{1}{x^2-x+1}$ and $\dfrac{x}{x^2-x+1}$ eventually after another substitution. However, it is a definite integral. It is easier to solve by the residue method than by partial fraction decomposition. Did you read the article I linked? It contains an example of how it is done, although it only mentions that $\int (x^3+1)^{-1} dx$ can be solved accordingly.

Allow me a remark: stop using $\infty$ as if it were a number, it is not! $\int_0^\infty \dfrac{1}{x+1}\,dx$ does not converge! This is another problem with the partial fraction decomposition: it leaves you with two divergent integrals! Since $\int_0^\infty (x^3+1)^{-1} dx$ is finite, the two "divergencies" cancel each other what cannot be seen by the partial fractions.

sorry i switch to the original question wihtout mentioning it

why say nasty anti-derivative? means i can't solve it?

i can't understand how you solve this
$\displaystyle \int \dfrac{x-2}{x^2-x+1}\, dx = \dfrac{1}{2} (\log(x^2 - x + 1) - 2 \sqrt{3} \tan^{-1}((2 x - 1)\sqrt{3})) + constant$

from where the log and tangent come? i think log only come with structure $\displaystyle \int \frac{1}{f(x)} \ dx$

can you please divide your steps into detailed steps?

i'll read your article and anything about residues after i solve the integration with partial fraction decomposition

 

[fresh_42]

sorry i switch to the original question wihtout mentioning it

why say nasty anti-derivative? means i can't solve it?

I find combinations with logarithms and arcus tangents nasty.

i can't understand how you solve this
$\displaystyle \int \dfrac{x-2}{x^2-x+1}\, dx = \dfrac{1}{2} (\log(x^2 - x + 1) - 2 \sqrt{3} \tan^{-1}((2 x - 1)\sqrt{3})) + constant$

I cheated.

integral ((x-2)/(x^2-x+1)) dx = - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

from where the log and tangent come? i think log only come with structure $\displaystyle \int \frac{1}{f(x)} \ dx$

can you please divide your steps into detailed steps?

I made no steps. I looked up the result.

i'll read your article and anything about residues after i solve the integration with partial fraction decomposition

You can't. Partial fraction decomposition gives you a problem you can't solve:

$$\dfrac{2\pi}{3\sqrt{3}}=\int_0^\infty \dfrac{1}{x^3+1}\,dx =\dfrac{1}{3}\underbrace{\int_0^\infty \dfrac{1}{x+1}\,dx}_{divergent}+\dfrac{2}{3}\underbrace{\int_0^\infty \dfrac{1}{x^2-x+1}\,dx}_{=\dfrac{4\pi}{3\sqrt{3}}}-\dfrac{1}{3}\underbrace{\int_0^\infty \dfrac{x}{x^2-x+1}\,dx}_{divergent}$$How will you deal with the divergent integrals? You cannot subtract one divergency from another and get zero. And even if, it won't produce the correct result. All you can do is solve the integral in the middle which gives you an arcus tangent, but this isn't the result of the entire integral.

Here are a couple of useful links where you can look up integrals:

https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integralrechnung
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integrale
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integraltransformationen

It's German and there is unfortunately no English version, but it is so full of formulas that you won't need a translation, and the technical terms are more or less identical. The only word you possibly need is "Beweis" which means "Proof".

 

[logistic_guy]

I cheated.

how teacher cheat?

You can't. Partial fraction decomposition gives you a problem you can't solve:

i'm very disappointed

why question mention partial fraction decomposition when this method is more difficult than residues?

How will you deal with the divergent integrals?

i don't know

Here are a couple of useful links where you can look up integrals:

https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integralrechnung
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integrale
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integraltransformationen

It's German and there is unfortunately no English version, but it is so full of formulas that you won't need a translation, and the technical terms are more or less identical. The only word you possibly need is "Beweis" which means "Proof".

i think i'll spend a couple of hours reading the german wiki before quiting the partial fraction decomposition

thank fresh_42

 

[fresh_42]

how teacher cheat?

It prevents me from talking nonsense

 

[mario99]

Because I am a fan of integration, I will continue solving the integral in details from where everyone got stuck. I will start from here:

$\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{3}\int_{0}^{\infty}\frac{x - 2}{x^2 - x + 1} \ dx$

We know that the derivative of $x^2 - x + 1 \ \text{is} \ \rightarrow 2x - 1$.

So, do this trick:

$\displaystyle x - 2 = \frac{1}{2}(2x - 1 - 3)$

Our integral will be:

$\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{3}\frac{1}{2}\int_{0}^{\infty}\frac{2x - 1 - 3}{x^2 - x + 1} \ dx$


$\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{-3}{x^2 - x + 1} \ dx$


$\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{1}{2}\int_{0}^{\infty}\frac{1}{x^2 - x + 1} \ dx$

We cannot factor $x^2 - x + 1 \rightarrow$ This gives us an idea to complete the square!

$\displaystyle x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + 1 - \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$

The main idea of completing the square is to have a form that looks like this: $\displaystyle \left( .... \right)^2 + 1$
So, we need to do some manipulation.

$\displaystyle \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} = \left(\frac{2x}{2} - \frac{1}{2}\right)^2 + \frac{3}{4} = \frac{1}{4}\left(2x - 1\right)^2 + \frac{3}{4} = \frac{3}{4}\frac{\left(2x - 1\right)^2}{3} + \frac{3}{4} = \frac{3}{4}\left[\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1\right]$

We have achieved what we wanted.

$\displaystyle \displaystyle x^2 - x + 1 = \frac{3}{4}\left[\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1\right]$

Our integral will be:

$\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{1}{2}\frac{4}{3}\int_{0}^{\infty}\frac{1}{\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1} \ dx$


$\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{2}{3}\int_{0}^{\infty}\frac{1}{\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1} \ dx$

Our integral is ready for cooking

At this stage beginner students will need to do $u$ substitution on paper. The substitution will give them this known form:

$A\displaystyle \int \frac{1}{u} \ du + B\int \frac{1}{u} \ du + C\int \frac{1}{u^2 + 1} \ du$

Where $A, B, \text{and} \ C$ are some constants.

Advanced students and I will compute the integral directly.

$\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{2}{3}\int_{0}^{\infty}\frac{1}{\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1} \ dx$


$\displaystyle = \frac{1}{3}\ln(x + 1)\bigg|_{0}^{\infty} - \frac{1}{6}\ln(x^2 - x + 1)\bigg|_{0}^{\infty} + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right)\bigg|_{0}^{\infty}$

We can combine the natural logarithmic functions together.

$\displaystyle = \frac{1}{6}\ln\left(\frac{(x + 1)^2}{x^2 - x + 1}\right)\bigg|_{0}^{\infty} + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right)\bigg|_{0}^{\infty}$

We will switch to limit notation.

$\displaystyle = \lim_{R\rightarrow \infty}\frac{1}{6}\ln\left(\frac{(x + 1)^2}{x^2 - x + 1}\right)\bigg|_{0}^{R} + \lim_{R\rightarrow \infty} \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right)\bigg|_{0}^{R}$


$\displaystyle = \lim_{R\rightarrow \infty}\frac{1}{6}\ln\left(\frac{R^2 + 2R + 1}{R^2 - R + 1}\right) - \frac{1}{6}\ln 1 + \lim_{R\rightarrow \infty} \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2R - 1}{\sqrt{3}}\right) - \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)$


$\displaystyle = \frac{1}{6}\ln 1 - \frac{1}{6}\ln 1 + \frac{1}{\sqrt{3}}\frac{\pi}{2} + \frac{1}{\sqrt{3}}\frac{\pi}{6}$


$\displaystyle = \frac{1}{\sqrt{3}}\frac{\pi}{2} + \frac{1}{\sqrt{3}}\frac{\pi}{6}$


$\displaystyle = \frac{3}{\sqrt{3}}\frac{\pi}{6} + \frac{1}{\sqrt{3}}\frac{\pi}{6}$


$\displaystyle = \frac{4}{\sqrt{3}}\frac{\pi}{6}$


$\displaystyle = \frac{2\pi}{3\sqrt{3}}$

 

[fresh_42]

Wow! And some nice tricks! I liked the step where you got rid of the two divergencies by using the properties of the logarithm!

My math teacher used to say: Anyone can differentiate, but integration is for artists!

 

[mario99]

Wow! And some nice tricks! I liked the step where you got rid of the two divergencies by using the properties of the logarithm!

Thanks fresh_42.

I am sure that you would have figured out all these tricks and manipulations if you took the integral seriously. But it seems that you and I are interested more in the complex analysis part because it is very rare to see a student discusses contours.

My math teacher used to say: Anyone can differentiate, but integration is for artists!

Believe it or not, a few weeks ago Mr. logistic_guy told us that he is an Artist. Therefore, I think that your teacher meant that integrals can only be solved by an Artist, but not all Artists can solve integrals.

The most fascinating thing here is that an Artist who studies engineering. A very rare thing to happen.

 

[khansaheb]

...... an Artist who studies engineering. A very rare thing to happen.

Sounds like you have not met Architectural Engineers.....

 

[mario99]

Sounds like you have not met Architectural Engineers.....

Like Albert Speer?

 

[khansaheb]

Like Albert Speer?

No.... not inhuman Nazi

I hoped you would know Frank Lloyd Wright or Charles-Édouard Jeanneret or Amenhotep..... oh how about Amenhotep or Michelangelo ....

 

[mario99]

No.... not inhuman Nazi

I hoped you would know Frank Lloyd Wright or Charles-Édouard Jeanneret or Amenhotep..... oh how about Amenhotep or Michelangelo ....

I have never heard of them. But why do you consider the architectural skills of Albert Speer as not an Art?

 

[khansaheb]

Albert Speer

Did you meet him.....

architectural skills of Albert Speer as not an Art?

Because he was an architect for Nazi camp !

 

[mario99]

I hoped you would know Frank Lloyd Wright or Charles-Édouard Jeanneret or Amenhotep..... oh how about Amenhotep or Michelangelo ....

Sorry, I just realized I have heard of Michelangelo in Ninja Turtles. (Orange eye mask.)

Did you meet him.....

Because he was an architect for Nazi camp !

Then, your post #29 was not written so precisely!

It should have been written like: Sounds like you have not met Architectural Engineers whom I want to choose.....

 

[logistic_guy]

Because I am a fan of integration, I will continue solving the integral in details from where everyone got stuck. I will start from here:

$\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{3}\int_{0}^{\infty}\frac{x - 2}{x^2 - x + 1} \ dx$

We know that the derivative of $x^2 - x + 1 \ \text{is} \ \rightarrow 2x - 1$.

So, do this trick:

$\displaystyle x - 2 = \frac{1}{2}(2x - 1 - 3)$

Our integral will be:

$\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{3}\frac{1}{2}\int_{0}^{\infty}\frac{2x - 1 - 3}{x^2 - x + 1} \ dx$


$\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{-3}{x^2 - x + 1} \ dx$


$\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{1}{2}\int_{0}^{\infty}\frac{1}{x^2 - x + 1} \ dx$

We cannot factor $x^2 - x + 1 \rightarrow$ This gives us an idea to complete the square!

$\displaystyle x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + 1 - \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$

The main idea of completing the square is to have a form that looks like this: $\displaystyle \left( .... \right)^2 + 1$
So, we need to do some manipulation.

$\displaystyle \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} = \left(\frac{2x}{2} - \frac{1}{2}\right)^2 + \frac{3}{4} = \frac{1}{4}\left(2x - 1\right)^2 + \frac{3}{4} = \frac{3}{4}\frac{\left(2x - 1\right)^2}{3} + \frac{3}{4} = \frac{3}{4}\left[\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1\right]$

We have achieved what we wanted.

$\displaystyle \displaystyle x^2 - x + 1 = \frac{3}{4}\left[\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1\right]$

Our integral will be:

$\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{1}{2}\frac{4}{3}\int_{0}^{\infty}\frac{1}{\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1} \ dx$


$\displaystyle = \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{2}{3}\int_{0}^{\infty}\frac{1}{\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1} \ dx$

Our integral is ready for cooking

At this stage beginner students will need to do $u$ substitution on paper. The substitution will give them this known form:

$A\displaystyle \int \frac{1}{u} \ du + B\int \frac{1}{u} \ du + C\int \frac{1}{u^2 + 1} \ du$

Where $A, B, \text{and} \ C$ are some constants.

Advanced students and I will compute the integral directly.

$\displaystyle \frac{1}{3}\int_{0}^{\infty}\frac{1}{x + 1} \ dx - \frac{1}{6}\int_{0}^{\infty}\frac{2x - 1}{x^2 - x + 1} \ dx + \frac{2}{3}\int_{0}^{\infty}\frac{1}{\left(\frac{2x - 1}{\sqrt{3}}\right)^2 + 1} \ dx$


$\displaystyle = \frac{1}{3}\ln(x + 1)\bigg|_{0}^{\infty} - \frac{1}{6}\ln(x^2 - x + 1)\bigg|_{0}^{\infty} + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right)\bigg|_{0}^{\infty}$

We can combine the natural logarithmic functions together.

$\displaystyle = \frac{1}{6}\ln\left(\frac{(x + 1)^2}{x^2 - x + 1}\right)\bigg|_{0}^{\infty} + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right)\bigg|_{0}^{\infty}$

We will switch to limit notation.

$\displaystyle = \lim_{R\rightarrow \infty}\frac{1}{6}\ln\left(\frac{(x + 1)^2}{x^2 - x + 1}\right)\bigg|_{0}^{R} + \lim_{R\rightarrow \infty} \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right)\bigg|_{0}^{R}$


$\displaystyle = \lim_{R\rightarrow \infty}\frac{1}{6}\ln\left(\frac{R^2 + 2R + 1}{R^2 - R + 1}\right) - \frac{1}{6}\ln 1 + \lim_{R\rightarrow \infty} \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2R - 1}{\sqrt{3}}\right) - \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)$


$\displaystyle = \frac{1}{6}\ln 1 - \frac{1}{6}\ln 1 + \frac{1}{\sqrt{3}}\frac{\pi}{2} + \frac{1}{\sqrt{3}}\frac{\pi}{6}$


$\displaystyle = \frac{1}{\sqrt{3}}\frac{\pi}{2} + \frac{1}{\sqrt{3}}\frac{\pi}{6}$


$\displaystyle = \frac{3}{\sqrt{3}}\frac{\pi}{6} + \frac{1}{\sqrt{3}}\frac{\pi}{6}$


$\displaystyle = \frac{4}{\sqrt{3}}\frac{\pi}{6}$


$\displaystyle = \frac{2\pi}{3\sqrt{3}}$

thank mario99 very much

i think i'll need to study your solution

hold on mario99, i'll get back to you soon

 

[logistic_guy]

i'm back

thank mario99

few remarks i don't understand

the completeing the square
how you jump from $\displaystyle x^2 - x + 1$ to $\displaystyle (x - \frac{1}{2})^2 + 1 - \frac{1}{4}$
it's not clear at all

how you decide this integration $\displaystyle \int_{0}^{\infty}\frac{1}{(\frac{x - 1}{\sqrt{3}})^2 + 1} \ dx = \frac{1}{\sqrt{3}}\tan^{-1}(\frac{x-1}{\sqrt{3}})$

i know tangent inverse is the angle inside a right triangle. i don't have triangle in the question

how this limit $\displaystyle \lim_{R\to\infty}\frac{1}{6}\ln(\frac{R^2 + 2R + 1}{R^2 - R + 1}) = \frac{1}{6}\ln 1$

how this limit $\displaystyle \lim_{R\to\infty}\tan^{-1}(\frac{2R - 1}{\sqrt{3}}) = \frac{\pi}{2}$

how $\displaystyle \tan^{-1}(\frac{-1}{\sqrt{3}}) = \frac{\pi}{6}$

you don't explain the steps of how you get them, so i got lost. i think i'm understand other calculations

 

[mario99]

the completeing the square
how you jump from $\displaystyle x^2 - x + 1$ to $\displaystyle (x - \frac{1}{2})^2 + 1 - \frac{1}{4}$
it's not clear at all

I have just added a quarter and subtracted a quarter.

$\displaystyle x^2 - x + 1 = x^2 - x + 1 + \frac{1}{4} - \frac{1}{4} \ \ \ \ \$ (I did not change anything.)

And from algebra class you should know: $\displaystyle x^2 - x + \frac{1}{4} = \left(x - \frac{1}{2}\right)\left(x - \frac{1}{2}\right) = \left(x - \frac{1}{2}\right)^2$.


Then, $\displaystyle x^2 - x + 1 = x^2 - x + 1 + \frac{1}{4} - \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 + 1 - \frac{1}{4}$

how you decide this integration $\displaystyle \int_{0}^{\infty}\frac{1}{(\frac{x - 1}{\sqrt{3}})^2 + 1} \ dx = \frac{1}{\sqrt{3}}\tan^{-1}(\frac{x-1}{\sqrt{3}})$

i know tangent inverse is the angle inside a right triangle. i don't have triangle in the question

You are right about the triangle, but we can also get the inverse tangent from other places. In the first day of integral calculus class, you should have been given the solution of some elementary integrals such as $\displaystyle \int \frac{1}{u^2 + 1} \ du = \tan^{-1} u + c.$
(Don't worry about the $c$, it is just some constant.)

If don't remember, you can follow this concept:

If the derivative of $x$ is $\rightarrow 1$, then the integral of $1$ is $\rightarrow x + c.$

This means:

If the derivative of $\tan^{-1}x$ is $\displaystyle \rightarrow \frac{1}{x^2 + 1}$, then the integral of $\displaystyle \frac{1}{x^2 + 1}$ is $\rightarrow \tan^{-1}x + c.$

Have you ever differentiated the inverse tangent function? Let me guess. No.

Then, it is time for differentiation

Let $y = \tan^{-1}x$

$\tan y = x$

$\displaystyle \frac{d}{dx}(\tan y = x)$

$\displaystyle \sec^2 y \ \frac{dy}{dx} = 1$

$\displaystyle \frac{dy}{dx} = \frac{1}{\sec^2 y}$

And we know from trigonometry class, $\tan^2 y + 1 = \sec^2 y$

Then,

$\displaystyle \frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{\tan^2 y + 1} = \frac{1}{x^2 + 1} \ \ \ \ \ \ \ \$ (And the proof is completed.)

how this limit $\displaystyle \lim_{R\to\infty}\frac{1}{6}\ln(\frac{R^2 + 2R + 1}{R^2 - R + 1}) = \frac{1}{6}\ln 1$

Limit involving infinity needs a lot of practice. But here we just have a fraction of polynomials. If the leading terms have the same power, the limit is just the coefficient of the leading terms division. This just means $\displaystyle \frac{R^2}{R^2} = \frac{1}{1} = 1$.

how this limit $\displaystyle \lim_{R\to\infty}\tan^{-1}(\frac{2R - 1}{\sqrt{3}}) = \frac{\pi}{2}$

Here we have $\tan^{-1}\infty$ and we want to know what happens to the inverse tangent function as $x$ goes to $\infty$. One way to see that $y$ goes to $\displaystyle \frac{\pi}{2}$ by graphing the function.

Another way is to write $\tan^{-1}\infty$ as:

$\displaystyle \tan y = \frac{\sin y}{\cos y} = \infty$

What is the first angle that will make the denominator $= 0$. Well, it is obvious $90 \ \text{degrees}$.

Then, $\displaystyle \tan \frac{\pi}{2} = \frac{\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}} = \frac{1}{0} = \infty$.

This means $\displaystyle \frac{\pi}{2} = \tan^{-1} \infty$.

Some people (such as fresh_42) will hate you if you write infinity $\infty$ as if it is a number. Therefore, it is a good practice to always use a limit:

$\displaystyle \lim_{x\rightarrow \infty}\tan^{-1}x = \frac{\pi}{2}$

how $\displaystyle \tan^{-1}(\frac{-1}{\sqrt{3}}) = \frac{\pi}{6}$

You mean $\displaystyle \tan^{-1} \frac{-1}{\sqrt{3}} = -\frac{\pi}{6}.$
This last one will depend on how quickly you can calculate some angles without calculator such as, $0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, \ \text{and} \ 90^{\circ}$.

Since the kindergarten, I have known:

$\displaystyle \sin 0^{\circ} = 0$
$\displaystyle \sin 30^{\circ} = \frac{1}{2}$
$\displaystyle \sin 45^{\circ} = \frac{\sqrt{2}}{2}$
$\displaystyle \sin 60^{\circ} = \frac{\sqrt{3}}{2}$
$\displaystyle \sin 90^{\circ} = 1$

And

$\displaystyle \cos 0^{\circ} = 1$
$\displaystyle \cos 30^{\circ} = \frac{\sqrt{3}}{2}$
$\displaystyle \cos 45^{\circ} = \frac{\sqrt{2}}{2}$
$\displaystyle \cos 60^{\circ} = \frac{1}{2}$
$\displaystyle \cos 90^{\circ} = 0$

I need a fraction from above that has $\displaystyle \sqrt{3}$, so I have $\displaystyle \sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\displaystyle \cos 30^{\circ} = \frac{\sqrt{3}}{2}$.

Let us try both:

$\displaystyle \tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$

This gives us: $\displaystyle 60^{\circ} = \tan^{-1} \sqrt{3} \ \ \ \ \$ (Not what I wanted.)

Let us try the second one:

$\displaystyle \tan 30^{\circ} = \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$

This gives us: $\displaystyle 30^{\circ} = \tan^{-1} \frac{1}{\sqrt{3}} \ \ \ \ \$ (Exactly what I wanted.)

This means: $\displaystyle \tan^{-1} \frac{-1}{\sqrt{3}} = -\tan^{-1} \frac{1}{\sqrt{3}} = -30^{\circ} = -\frac{\pi}{6}\ \ \ \ \$

 

[logistic_guy]

I have just added a quarter and subtracted a quarter.

$\displaystyle x^2 - x + 1 = x^2 - x + 1 + \frac{1}{4} - \frac{1}{4} \ \ \ \ \$ (I did not change anything.)

And from algebra class you should know: $\displaystyle x^2 - x + \frac{1}{4} = \left(x - \frac{1}{2}\right)\left(x - \frac{1}{2}\right) = \left(x - \frac{1}{2}\right)^2$.


Then, $\displaystyle x^2 - x + 1 = x^2 - x + 1 + \frac{1}{4} - \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 + 1 - \frac{1}{4}$


You are right about the triangle, but we can also get the inverse tangent from other places. In the first day of integral calculus class, you should have been given the solution of some elementary integrals such as $\displaystyle \int \frac{1}{u^2 + 1} \ du = \tan^{-1} u + c.$
(Don't worry about the $c$, it is just some constant.)

If don't remember, you can follow this concept:

If the derivative of $x$ is $\rightarrow 1$, then the integral of $1$ is $\rightarrow x + c.$

This means:

If the derivative of $\tan^{-1}x$ is $\displaystyle \rightarrow \frac{1}{x^2 + 1}$, then the integral of $\displaystyle \frac{1}{x^2 + 1}$ is $\rightarrow \tan^{-1}x + c.$

Have you ever differentiated the inverse tangent function? Let me guess. No.

Then, it is time for differentiation

Let $y = \tan^{-1}x$

$\tan y = x$

$\displaystyle \frac{d}{dx}(\tan y = x)$

$\displaystyle \sec^2 y \ \frac{dy}{dx} = 1$

$\displaystyle \frac{dy}{dx} = \frac{1}{\sec^2 y}$

And we know from trigonometry class, $\tan^2 y + 1 = \sec^2 y$

Then,

$\displaystyle \frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{\tan^2 y + 1} = \frac{1}{x^2 + 1} \ \ \ \ \ \ \ \$ (And the proof is completed.)


Limit involving infinity needs a lot of practice. But here we just have a fraction of polynomials. If the leading terms have the same power, the limit is just the coefficient of the leading terms division. This just means $\displaystyle \frac{R^2}{R^2} = \frac{1}{1} = 1$.


Here we have $\tan^{-1}\infty$ and we want to know what happens to the inverse tangent function as $x$ goes to $\infty$. One way to see that $y$ goes to $\displaystyle \frac{\pi}{2}$ by graphing the function.

Another way is to write $\tan^{-1}\infty$ as:

$\displaystyle \tan y = \frac{\sin y}{\cos y} = \infty$

What is the first angle that will make the denominator $= 0$. Well, it is obvious $90 \ \text{degrees}$.

Then, $\displaystyle \tan \frac{\pi}{2} = \frac{\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}} = \frac{1}{0} = \infty$.

This means $\displaystyle \frac{\pi}{2} = \tan^{-1} \infty$.

Some people (such as fresh_42) will hate you if you write infinity $\infty$ as if it is a number. Therefore, it is a good practice to always use a limit:

$\displaystyle \lim_{x\rightarrow \infty}\tan^{-1}x = \frac{\pi}{2}$


You mean $\displaystyle \tan^{-1} \frac{-1}{\sqrt{3}} = -\frac{\pi}{6}.$
This last one will depend on how quickly you can calculate some angles without calculator such as, $0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, \ \text{and} \ 90^{\circ}$.

Since the kindergarten, I have known:

$\displaystyle \sin 0^{\circ} = 0$
$\displaystyle \sin 30^{\circ} = \frac{1}{2}$
$\displaystyle \sin 45^{\circ} = \frac{\sqrt{2}}{2}$
$\displaystyle \sin 60^{\circ} = \frac{\sqrt{3}}{2}$
$\displaystyle \sin 90^{\circ} = 1$

And

$\displaystyle \cos 0^{\circ} = 1$
$\displaystyle \cos 30^{\circ} = \frac{\sqrt{3}}{2}$
$\displaystyle \cos 45^{\circ} = \frac{\sqrt{2}}{2}$
$\displaystyle \cos 60^{\circ} = \frac{1}{2}$
$\displaystyle \cos 90^{\circ} = 0$

I need a fraction from above that has $\displaystyle \sqrt{3}$, so I have $\displaystyle \sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\displaystyle \cos 30^{\circ} = \frac{\sqrt{3}}{2}$.

Let us try both:

$\displaystyle \tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$

This gives us: $\displaystyle 60^{\circ} = \tan^{-1} \sqrt{3} \ \ \ \ \$ (Not what I wanted.)

Let us try the second one:

$\displaystyle \tan 30^{\circ} = \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$

This gives us: $\displaystyle 30^{\circ} = \tan^{-1} \frac{1}{\sqrt{3}} \ \ \ \ \$ (Exactly what I wanted.)

This means: $\displaystyle \tan^{-1} \frac{-1}{\sqrt{3}} = -\tan^{-1} \frac{1}{\sqrt{3}} = -30^{\circ} = -\frac{\pi}{6}\ \ \ \ \$

thank mario99

i think i'm understanding

My point was if you write the fraction as $\displaystyle \frac{1}{6}$ when it is meant to be $\displaystyle \frac{5}{6}$, you are missing some terms in the numerator. These terms are not difficult to find when you are calculating simple fractions as $\displaystyle \frac{1}{2} + \frac{1}{3}$. But when the fractions are polynomials, you will need to do some work to find the missing terms as was explained to you in post #18 by fresh_42.

As professor Dave has told you, it is good to know how to solve by partial fraction, but in this problem you have to put this method aside as you were told not to use it. If I were you, I would directly solve the problem by complex analysis.


Thank you for uploading this picture. It helps a lot to understand what to calculate. In the contour, you have three paths, so you will need to solve three integrals:

$\displaystyle \int_{C_1} f(z) \ dz + \int_{C_2} f(z) \ dz + \int_{C_3} f(z) \ dz = 2\pi iB$

If you have solved contours before, you would already know that the curve path would get cancelled as it would be equal to zero. Therefore, two paths would remain:

$\displaystyle \int_{C_1} f(z) \ dz + \int_{C_3} f(z) \ dz = 2\pi iB$

Your first task is to find $\displaystyle B$ which is the residues at the zeros that are located inside the contour. To find $\displaystyle B$, you need first to find the zeros which is the easiest part in the problem.

$\displaystyle x^3 + 1 = 0$ (Since this is a cubic function, you should get three zeros. Some are imaginary.)

I remember that there is a post written by fresh_42 where he showed the method of finding all the roots (real and complex) of a function. You can use it as a quick guidance. If you got stuck in finding the complex roots, use W|A.

Note: It is not difficult to show that the integral of the curve path will be zero as $\displaystyle R \rightarrow \infty.$

$\displaystyle x^3 + 1 = 0$
$\displaystyle x^3 = -1$
$\displaystyle x = \sqrt[3]{-1}$
$\displaystyle \sqrt[3]{-1} = \sqrt[3]{1}e^{\frac{\pi i + 2k\pi i}{3}}$
$\displaystyle x_1 = e^{\frac{\pi i}{3}} = \cos \frac{\pi}{3} + i\sin \frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$
$\displaystyle x_2 = e^{\pi i} = \cos \pi + i\sin \pi = -1$
$\displaystyle x_3 = e^{\frac{5\pi i}{3}} = \cos \frac{5\pi}{3} + i\sin \frac{5\pi}{3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$

i find the roots

Integration is a complicated subject. There are many tips and tricks but there is no golden way to solve them all.
Here is an overview of the residue method:

An Overview of Complex Differentiation and Integration

I want to shed some light on complex analysis without getting all the technical details in the way which are necessary for the precise treatments that can be found in many excellent standard textbooks.

www.physicsforums.com

thank fresh_42

i read the site. it's very complicated

 

[mario99]

$\displaystyle x_1 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$
$\displaystyle x_2 = -1$
$\displaystyle x_3 = \frac{1}{2} - i\frac{\sqrt{3}}{2}$

i find the roots

Very good. That was super. Now which of these zeros lie on or in the contour? Think carefully.

 

[fresh_42]

thank fresh_42

i read the site. it's very complicated

Yes, sorry, @logistic_guy . It is a bit hard to tell which level of answers should be given. I don't know much about the members. Some ask very basic questions, others quite sophisticated ones, and yours swam somewhere between them. Complex integration of real functions is not trivial. And real integration of real functions can be tricky as @mario99 has shown us.

I'm still baffled by how he got rid of the divergence problem of the two critical terms before integration. It meant that he used partial fractions but took a step back and recombined two critical quotients again to avoid infinite values. I very much like that example because it shows that "one method serves all" usually doesn't work for integrations.

I don't want to bother you with another article, but I quote it anyway in case a) some readers might be interested, b) you're brave and give me another shot (I first thought your question was primarily about complex integration), c) someone wants to link it because it lists many of the standard tricks in real integration. And if nothing of it counts: it has examples!
www.physicsforums.com/insights/the-art-of-integration/

Don't give up. Integration can be fun:

Because I am a fan of integration, I will continue solving the integral in details from where everyone got stuck.