If Du is Smaller Then Dx

[Jason76]

\(\displaystyle \int \dfrac{2x}{x + 1}dx\)

\(\displaystyle u = (x + 1)\)

\(\displaystyle du = 1\)

\(\displaystyle 2 \int \cfrac{\cfrac{2x}{2}dx}{x + 1}\) Divide by dx by 1/2 to make du = dx. Balance the equation with 2 (the reciprocal of 1/2).

\(\displaystyle 2 \int \cfrac{\cfrac{2x}{2}dx}{x + 1} = (2)\ln(x + 1) + C\) (Final Answer)

But the book came up with a different by almost similar answer. Didn't they try to balance functions (ex:2x) which is illegal?

\(\displaystyle \int \dfrac{2x}{x + 1}dx = 2x - 2 \ln(x + 1) + C\)

 

[soroban]

Hello, Jason76!

Your substitution is incorrect.

\(\displaystyle \displaystyle\int \frac{2x}{x + 1}dx\)

Long division: .\(\displaystyle \dfrac{2x}{x+1} \;=\;2 - \dfrac {2}{x+1}\)

Then: .\(\displaystyle \displaystyle \int\left(2 - \frac{2}{x+1}\right)dx \;=\;2x - 2\ln|x+1| + C\)

 

[Jason76]

Hello, Jason76!

Your substitution is incorrect.


Long division: .\(\displaystyle \dfrac{2x}{x+1} \;=\;2 - \dfrac {2}{x+1}\)

Then: .\(\displaystyle \displaystyle \int\left(2 - \frac{2}{x+1}\right)dx \;=\;2x - 2\ln|x+1| + C\)

So when du is less than dx, then long division is the way to go.

 

[HallsofIvy]

\(\displaystyle \int \dfrac{2x}{x + 1}dx\)

\(\displaystyle u = (x + 1)\)

\(\displaystyle du = 1\)

you mean du= dx, I hope.

\(\displaystyle 2 \int \cfrac{\cfrac{2x}{2}dx}{x + 1}\) Divide by dx by 1/2 to make du = dx. Balance the equation with 2 (the reciprocal of 1/2)

du is equal to dx! Just take the 2 out of the integral:
\(\displaystyle \int \frac{2x}{x+1}dx= 2\int \frac{x}{x+ 1} dx\)
Letting u= x+ 1, du= dx and x=u- 1 so the integral becomes
\(\displaystyle 2\int \frac{u- 1}{u}du= 2\int( 1- \frac{1}{u})du\)

\(\displaystyle 2 \int \cfrac{\cfrac{2x}{2}dx}{x + 1} = (2)\ln(x + 1) + C\) (Final Answer)

I don't see how you jumped from the integral to the "answer". And what happened to u?

But the book came up with a different by almost similar answer. Didn't they try to balance functions (ex:2x) which is illegal?

\(\displaystyle \int \dfrac{2x}{x + 1}dx = 2x - 2 \ln(x + 1) + C\)

 

[MarkFL]

Personally I would write:

\(\displaystyle \displaystyle \frac{2x}{x+1}\,dx=2\int\frac{(x+1)-1}{x+1}\,dx=2\int 1-\frac{1}{x+1}\,dx=2(x-\ln|x+1|)+C\)

However, if you wish to use a substitution you could let:

\(\displaystyle u=x+1\,\therefore\,du=dx\) and you have:

\(\displaystyle \displaystyle 2\int\frac{u-1}{u}\,dx=2\int1-\frac{1}{u}\,du=2(u-\ln|u|)+C=2((x+1)-\ln|x+1|)+C=2(x-\ln|x+1|)+C\)

Note: the constant term 2 was "absorbed" by the constant of integration in the last step.

 

[Jason76]

Du Smaller than Dx Problem

\(\displaystyle \int \dfrac{2x}{x + 1}dx\)

\(\displaystyle u = (x + 1)\)

\(\displaystyle du = 1\)

Reasoning Behind the Problem (according to the book):

\(\displaystyle \int \dfrac{2x}{x + 1}dx = 2 \int \dfrac{x}{x + 1}dx\)

\(\displaystyle = 2 \int (\dfrac{x + 1}{x + 1} - \dfrac{1}{x + 1})dx\)

\(\displaystyle = 2 \int (1 - \dfrac{1}{x + 1})dx\)

\(\displaystyle = 2 \int dx - 2 \int \dfrac{dx}{x + 1}\)

\(\displaystyle \int \dfrac{2x}{x + 1}dx = 2x - 2 \ln(x + 1) + C\) Answer

But have no clue what's going on (in any of it). Any hints?

Of course I understand what to do if du = dx at the start. In that case, you do nothing. However, if du is bigger than dx than you multiply the dx by a constant, and then balance it with the reciprocal (of the constant) in the answer.

 

[daon2]

Jason, your question was answered in detail more than once above. You seem to be stuck with the idea that substitution is necessary right off the bat. It is not. Algebra was done to simplify the integrand before any "calculus" was done.

In short, when attempting to integrate a rational function where the the degree of the numerator is bigger than or equal to the degree of the denominator, long division as a first step is usually the way to go.

 

[Jason76]

Jason, your question was answered in detail more than once above. You seem to be stuck with the idea that substitution is necessary right off the bat. It is not. Algebra was done to simplify the integrand before any "calculus" was done.

In short, when attempting to integrate a rational function where the the degree of the numerator is bigger than or equal to the degree of the denominator, long division as a first step is usually the way to go.

Ok, so the first step is to divide 2x by x + 1 using long division. Ok I will work on it, and get back to you.

 

[daon2]

Ok, so the first step is to divide 2x by x + 1 using long division.

That is one way to go, sure. The book opted to take the factor of 2 out first. While not necessary, it will make the rest of your work easier.

 

[Jason76]

That is one way to go, sure. The book opted to take the factor of 2 out first. While not necessary, it will make the rest of your work easier.

Actually, with long division I already got the \(\displaystyle 2x - 2\) for the final answer (part of the final answer), right off the bat. With such an easy shortcut, why does the book have all that in-between work?

 

[daon2]

Actually, with long division I already got the \(\displaystyle 2x - 2\) for the final answer (part of the final answer), right off the bat. With such an easy shortcut, why does the book have all that in-between work?

I'm not sure I understand. Removing the factor of two simplifies the integral.

\(\displaystyle \displaystyle \int \dfrac{x}{x+1} dx\)

Is (minimally) an "easier" integral than

\(\displaystyle \displaystyle \int \dfrac{2x}{x+1} dx=\int (2) \dfrac{x}{x+1} dx\)

So bringing the two outside makes easier the "calculus part" of the problem, which is really what you should be concerned with.

For instance: \(\displaystyle \int \dfrac{7\pi x}{2x+2}dx\) is of the same difficulty, since:

\(\displaystyle \displaystyle \int \dfrac{7\pi x}{2x+2}dx = \int \left(\dfrac{7\pi }{2}\right)\dfrac{x}{x+1}dx= \left(\dfrac{7\pi}{2}\right)\int \dfrac{x}{x+1}dx\).

So calculate the simpler integral, and multiply by the constant you factored out of the integral to get your final answer.

 

[Jason76]

I'm not sure I understand. Removing the factor of two simplifies the integral.

\(\displaystyle \displaystyle \int \dfrac{x}{x+1} dx\)

Is (minimally) an "easier" integral than

\(\displaystyle \displaystyle \int \dfrac{2x}{x+1} dx=\int (2) \dfrac{x}{x+1} dx\)

So bringing the two outside makes easier the "calculus part" of the problem, which is really what you should be concerned with.

For instance: \(\displaystyle \int \dfrac{7\pi x}{2x+2}dx\) is of the same difficulty, since:

\(\displaystyle \displaystyle \int \dfrac{7\pi x}{2x+2}dx = \int \left(\dfrac{7\pi }{2}\right)\dfrac{x}{x+1}dx= \left(\dfrac{7\pi}{2}\right)\int \dfrac{x}{x+1}dx\).

So calculate the simpler integral, and multiply by the constant you factored out of the integral to get your final answer.

I originally tried to do "regular division" on the numerator (of the original problem), because I recognized the goal was to reduce 2x to x. However, this method doesn't lead to the final answer in the book, so I tried "long division". So I did long division, but I can't relate what's done with long division, with the actual problem. Where in the book's version of the "reasoning behind the problem" is the long division implemented?

 

[daon2]

I originally tried to do "regular division" on the numerator (of the original problem), because I recognized the goal was to reduce 2x to x. However, this method doesn't lead to the final answer in the book, so I tried "long division". So I did long division, but I can't relate what's done with long division, with the actual problem. Where in the book's version of the "reasoning behind the problem" is the long division implemented?

The book's solution used an alternate form of long division.

In the straight-forward version:

\(\displaystyle \hspace{1.5cm}1\\x+1\,\,\,\overline{)\,\, x\,\,\,\,\,\,\,\,\,\,\,\,\,}\hspace{0.5cm} \implies \hspace{0.5cm} \dfrac{x}{x+1} = 1 + \dfrac{-1}{x+1} = 1 - \dfrac{1}{x+1}\\ \hspace{0.8cm} \underline{-(\,\, x+1)\,\,\,} \\ \hspace{1.6cm}\, - 1 \,\, \text{( = remainder)}\)

In the book's version (also what has been posted by others), the author saw that by adding 1 to the numerator, it was the same as the denominator, hence they would cancel. To keep the integrad the same, though, you must also subtract 1 from the numerator. So you;re really adding 0 in a fancy way. So:

\(\displaystyle \dfrac{x}{x+1} = \dfrac{x + \overbrace{ 1 - 1}^{\text{adding 0}}}{x+1} = \dfrac{x+1}{x+1} - \dfrac{1}{x+1} = 1-\dfrac{1}{x+1}\)

 

[JeffM]

I originally tried to do "regular division" on the numerator (of the original problem), because I recognized the goal was to reduce 2x to x. However, this method doesn't lead to the final answer in the book, so I tried "long division". So I did long division, but I can't relate what's done with long division, with the actual problem. Where in the book's version of the "reasoning behind the problem" is the long division implemented?

Jason, as usual, you are trying to formulate mechanical rules where either there are none or there are some but they differ from what you formulate.

\(\displaystyle \int k * f(x)\ dx = k * \int f(x)\ dx.\) Integrating a constant times a function is the same as multiplying the constant times the integral of the function.

\(\displaystyle \int \dfrac{2x}{x + 1}\ dx = 2 * \int \dfrac{x}{x + 1}\ dx\) by the rule above, and division has nothing to do with it.

I have no clue what you talking about with du and dx. What substitution are you trying? You do not specify one. The book does not use a substitution.

If I do algebraic long division, I get: \(\displaystyle \dfrac{x}{x + 1} = 1 - \dfrac{1}{x + 1}.\) If you don't believe me, multiply both sides by x + 1.

The book gets the exact same answer a different way

\(\displaystyle \dfrac{x}{x + 1} = \dfrac{x + 1 - 1}{x + 1} = \dfrac{x + 1}{x + 1} - \dfrac{1}{x + 1} = 1 - \dfrac{1}{x + 1}.\) Avoids long division.

Here is another rule

\(\displaystyle f(x) = g(x) \pm h(x) \implies \int f(x)\ dx = \int \{g(x) \pm h(x)\}\ dx = \int g(x)\ dx \pm \int h(x)\ dx.\)

So \(\displaystyle \int \dfrac{2x}{x + 1}\ dx = 2 \int \dfrac{x}{x + 1}\ dx = 2\int \left\{1 - \dfrac{1}{x + 1}\right\}\ dx = 2\left\{\int {dx} - \int \dfrac{1}{x + 1}\ dx\right\}.\)

Now to solve the second integral, you CAN use a substitution:

\(\displaystyle u = x + 1 \implies \dfrac{du}{dx} = 1 \implies du = dx \implies \int \dfrac{1}{u}\ du = ln|u| + K.\)

So du is NOT less than dx.

And \(\displaystyle \int \dfrac{2x}{x + 1}\ dx = 2\left\{\int {dx} - \int \dfrac{1}{u}\ du\right\} = 2(x + M - ln|u| + K) = 2x - 2ln|x + 1| + C.\)

 

[lookagain]

Personally I would write:

\(\displaystyle \displaystyle \frac{2x}{x+1}\,dx = \) . . . You're missing the integration symbol at the outset.


\(\displaystyle 2\int\frac{(x+1)-1}{x+1}\,dx\)

...

 

[Jason76]

The book's solution used an alternate form of long division.

In the straight-forward version:

\(\displaystyle \hspace{1.5cm}1\\x+1\,\,\,\overline{)\,\, x\,\,\,\,\,\,\,\,\,\,\,\,\,}\hspace{0.5cm} \implies \hspace{0.5cm} \dfrac{x}{x+1} = 1 + \dfrac{-1}{x+1} = 1 - \dfrac{1}{x+1}\\ \hspace{0.8cm} \underline{-(\,\, x+1)\,\,\,} \\ \hspace{1.6cm}\, - 1 \,\, \text{( = remainder)}\)

In the book's version (also what has been posted by others), the author saw that by adding 1 to the numerator, it was the same as the denominator, hence they would cancel. To keep the integrad the same, though, you must also subtract 1 from the numerator. So you;re really adding 0 in a fancy way. So:

\(\displaystyle \dfrac{x}{x+1} = \dfrac{x + \overbrace{ 1 - 1}^{\text{adding 0}}}{x+1} = \dfrac{x+1}{x+1} - \dfrac{1}{x+1} = 1-\dfrac{1}{x+1}\)

Ok, got it so far, but how do you get from \(\displaystyle 1 - \dfrac{1}{x + 1}\) to \(\displaystyle 2x - 2\) (in the final answer) ? The whole answer being \(\displaystyle 2x - 2 \ln(x + 1) + C\)

 

[JeffM]

Ok, got it so far, but how do you get from \(\displaystyle 1 - \dfrac{1}{x + 1}\) to \(\displaystyle 2x - 2\) (in the final answer) ? The whole answer being \(\displaystyle 2x - 2 \ln(x + 1) + C\)

\(\displaystyle \int \dfrac{2x}{x + 1}\ dx = 2 * \int \dfrac{x}{x + 1}\ dx = 2 * \int \left(1 - \dfrac{1}{x + 1}\right)\ dx = 2\left\{\int dx - \int\dfrac{1}{x + 1}\ dx\right\} = 2(x + M - ln|x + 1| + K) = 2x - 2ln|x + 1| + C.\)

 

[Jason76]

\(\displaystyle \int \dfrac{2x}{x + 1}\ dx = 2 * \int \dfrac{x}{x + 1}\ dx = 2 * \int \left(1 - \dfrac{1}{x + 1}\right)\ dx = 2\left\{\int dx - \int\dfrac{1}{x + 1}\ dx\right\} = 2(x + M - ln|x + 1| + K) = 2x - 2ln|x + 1| + C.\)

What are M and K?

 

[JeffM]

What are M and K?

When I integrate \(\displaystyle \int\ dx\), I get x + a constant, which I have called M.

When I integrate \(\displaystyle \int \dfrac{1}{x + 1}\ dx\), I get ln|x + 1| + a constant, which I have called K.

Let 2M + 2K = C.

Then 2 * (x + M + ln|x + 1| + K) = 2x + 2ln|x + 1| + 2M + 2K = 2x + 2ln|x + 1| + C.

Every indefinite integral resolves into a function plus a constant, right?

 

[Jason76]

When I integrate \(\displaystyle \int\ dx\), I get x + a constant, which I have called M.

When I integrate \(\displaystyle \int \dfrac{1}{x + 1}\ dx\), I get ln|x + 1| + a constant, which I have called K.

Let 2M + 2K = C.

Then 2 * (x + M + ln|x + 1| + K) = 2x + 2ln|x + 1| + 2M + 2K = 2x + 2ln|x + 1| + C.

Every indefinite integral resolves into a function plus a constant, right?

\(\displaystyle \int \dfrac{2x}{x + 1}\ dx = 2 * \int \dfrac{x}{x + 1}\ dx = 2 * \int \left(1 - \dfrac{1}{x + 1}\right)\ dx = 2\left\{\int dx - \int\dfrac{1}{x + 1}\ dx\right\} = 2(x + M - ln|x + 1| + K) = 2x - 2ln|x + 1| + C.\)

Ok, I understand most of it, but how did the 1 (on the left) on \(\displaystyle (1 - \dfrac{1}{1 + x})\) disappear, and why is there a negative sign next to the ln at some point in the equation?

Steps for Integration if it's in Fractional Form and the Du is smaller than Dx

Ok so here are some steps summed up:

1. Seperate the single constant from the dx. (They are next to each other.)

2. Do long division.

3. Simplify the long division answer. (if needed)

4. Integrate the dx and the simplified long division answer. Note: Integrating the dx leads to an "x and an M constant" and integrating the long divison answer leads to the integration formula \(\displaystyle \ln|u| + C\)

5. Multiply single constant by the "x" and the "M constant" (where "x and M" both individually equal one) and \(\displaystyle \ln|u| + C\) integration formula.

6. You have the final answer