Integratin problem

[hank]

S x^2 / (x^2 - 3x + 2) dx Limits are 1 to 3/2

After doing the integration, I get:

lim b->1+ [x - ln(x - 1) + 4ln(x - 2)] limits b to 3/2

Which I evaluate to:

4ln(1/2) - ln(1/2) + (1/2) - (-infinity)

Which I evaluate to infinity, and therefore it diverges.


However, this answer is apparently wrong.

I understand the correct answer to be :

ln(1/2) - (-infinity) and it therefore diverges.

Can someone confirm if I'm right or wrong, and which part I'm wrong on?

 

[arthur ohlsten]

x^2/[x^2-3x+2] improper fraction divide

1 + [3x-2]/[x^2-3x+2] factor
1 + [3x-2] /{ [x-2][x-1]

let [3x-2] /[x-2][x-1] = A/[x-2] + B/[x-1]
[3x-2]=A[x-1]+B[x-2]
3x=[A+B]x
-2=-A-2B

3=A+B
-2=-A-2B add equations
1=-B then
A=4

rewriting the integral
S[1,3/2] x^2/[x^2-3x+2]= S 1dx +S 4dx/[x-2] -S dx/[x-1]
integral= x +4ln[x-2] -ln [x-1]
integral = x + ln [x-2]^4/[x-1] evaluated at 3/2 and 1
integral= [3/2-1] +ln{[-1/2]^4/1/2} - ln {[-1]^4/0
integral= 1/2 +1/8 -infinity answer

please check math
Arthur

 

[soroban]

Hello, hank!

Here's an alternative approach . . .
It is as long as (if not longer than) using Partial Fractions,
. . but it is a handy method to keep in your arsenal.
It also teaches you Olympic-level gymnastics.

\(\displaystyle \L\int^{\;\;\;3/2}_1\frac{x^2}{x^2\,-\,3x\,+\,2}\,dx\)

As Arthur suggested, long division: \(\displaystyle \L\:\frac{x^2}{x^2\,-\,3x\,+\,2}\:=\:1\,+\,\frac{3x\,-\,2}{x^2\,-\,3x\,+\,2}\)

The fraction can be integrated if we hammer at it long enough . . .


Multiply the numerator by \(\displaystyle \L\frac{3}{2}\cdot\frac{2}{3}:\;\;\frac{3}{2}\cdot\frac{2}{3}(3x\,-\,2) \:=\:\frac{3}{2}\left(2x\,-\,\frac{4}{3}\right)\)


Inside the parentheses, subtract 3 and add 3:
. . \(\displaystyle \L\:\frac{3}{2}\bigg(2x\,\)- 3 \(\displaystyle \L\,-\,\frac{4}{3}\,\) + 3\(\displaystyle \L\bigg) \;=\;\frac{3}{2}\bigg(2x\,-\,3\,+\,\frac{5}{3}\bigg)\;=\;\frac{3}{2}(2x\,-\,3)\,+\,\frac{5}{2}\)


The fraction becomes: \(\displaystyle \L\:\frac{\frac{3}{2}(2x\,-\,3)\,+\,\frac{5}{2}}{x^2\,-\,3x\,+\,2} \;=\;\frac{3}{2}\cdot\frac{2x\,-\,3}{x^2\,-\,3x\,+\,2}\,+\,\frac{5}{2}\cdot\frac{1}{x^2\,-\,3x\,+\,2}\)

In the second denominator, complete the square:
. . \(\displaystyle \L x^2\,-\,3x\,+\,2 \;=\;x^2\,-\,3x\,+\,\frac{9}{4}\,+\,2\,-\,\frac{9}{4} \;=\;\left(x\,-\,\frac{3}{2}\right)^2\,-\,\frac{1}{4}\)

. . and we have: \(\displaystyle \L\:1\,+\,\frac{3}{2}\cdot\frac{2x\,-\,3}{x^2\,-\,3x\,+\,2}\,+\,\frac{5}{2}\cdot\frac{1}{(x\,-\,\frac{3}{2})^2 - \frac{1}{4}}\)


Integrate: \(\displaystyle \L\:\int 1\,dx \;+\;\frac{3}{2}\int\frac{2x\,-\,3}{x^2\,-\,3x\,+\,2} \;+\;\frac{5}{2}\int\frac{dx}{(x-\frac{3}{2})^2\,+\,\frac{1}{4}}\)

. . In the second integral, substitute: \(\displaystyle u\:=\:x^2\,-\,3x\,+\,2\)

. . In the third integral, substitute: \(\displaystyle v\:=\:x\,-\,\frac{3}{2}\)

\(\displaystyle \text{And we have: }\L\:\int1\,dx\;+\;\frac{3}{2}\underbrace{\int\frac{du}{u}}_{\ln}\,+\,\frac{5}{2}\underbrace{\int\frac{dv}{v^2\,+\,\frac{1}{4}}}_{\arctan}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Yes, I agree . . . it requires a lot of steps.
. . But the steps are from Algebra I only.

This technique is used when a quadratic denominator does not factor.

It is also a method guarenteed to impress/surprise/terrify your teacher.

 

[galactus]

Soroban, ol' buddy, you're showin' off. But in a good way

 

[arthur ohlsten]

I started to do the problem the way Soroban did it.
I would have continued, but it was obvious that the approach was involved, and I doubted the instructor meant the problem to be attacked in that way.
I assumed you were studying the method I used

I find factoring the denominator an easier approach, and I would use it when teaching L'Place Transforms.

A quadratic can always be factored, and will contain "imaginary" roots.
The fraction can then be 'broken" into two separate fractions
Arthur

 

[Count Iblis]

I started to do the problem the way Soroban did it.
I would have continued, but it was obvious that the approach was involved, and I doubted the instructor meant the problem to be attacked in that way.
I assumed you were studying the method I used

I find factoring the denominator an easier approach, and I would use it when teaching L'Place Transforms.

A quadratic can always be factored, and will contain "imaginary" roots.
The fraction can then be 'broken" into two separate fractions
Arthur

I agree. And partial fractions can be obtained very fast by just expanding around every pole, adding up all the terms with negative exponents. To see this take the difference of the original fraction (with degree of numerator less than that of denominator) and the Laurent expansion terms. The function no longer has any poles but is bounded so, by Liouville's theorem, must be a constant (which is zero).

 

[Count Iblis]

Let's illustrate partial fractions using complex analyses.

\(\displaystyle f(x)=\frac{x^2}{x^2 - 3x + 2}\)

\(\displaystyle f(x)\) has simple poles at \(\displaystyle x=1\) and \(\displaystyle x=2\). The residues at the poles is easily found using L'hopital's rule:

\(\displaystyle \lim_{x\rightarrow 1}(x-1)f(x)=-1\)

\(\displaystyle \lim_{x\rightarrow 2}(x-2)f(x)=4\)

Consider the function:

\(\displaystyle s(x) = \frac{-1}{x-1}+ \frac{4}{x-2}\)

Then \(\displaystyle f(x)-s(x)\) has no poles in the complex plane, but it is clear that this function is bounded so it must be a constant. Since the limit to infinity is 1, the constant is 1. So we have:

\(\displaystyle f(x) = 1 + s(x)\)

 

[Count Iblis]

Hello, hank!

Here's an alternative approach . . .
It is as long as (if not longer than) using Partial Fractions,
. . but it is a handy method to keep in your arsenal.
It also teaches you Olympic-level gymnastics.

\(\displaystyle \L\int^{\;\;\;3/2}_1\frac{x^2}{x^2\,-\,3x\,+\,2}\,dx\)

As Arthur suggested, long division: \(\displaystyle \L\:\frac{x^2}{x^2\,-\,3x\,+\,2}\:=\:1\,+\,\frac{3x\,-\,2}{x^2\,-\,3x\,+\,2}\)

The fraction can be integrated if we hammer at it long enough . . .


Multiply the numerator by \(\displaystyle \L\frac{3}{2}\cdot\frac{2}{3}:\;\;\frac{3}{2}\cdot\frac{2}{3}(3x\,-\,2) \:=\:\frac{3}{2}\left(2x\,-\,\frac{4}{3}\right)\)


Inside the parentheses, subtract 3 and add 3:
. . \(\displaystyle \L\:\frac{3}{2}\bigg(2x\,\)- 3 \(\displaystyle \L\,-\,\frac{4}{3}\,\) + 3\(\displaystyle \L\bigg) \;=\;\frac{3}{2}\bigg(2x\,-\,3\,+\,\frac{5}{3}\bigg)\;=\;\frac{3}{2}(2x\,-\,3)\,+\,\frac{5}{2}\)


The fraction becomes: \(\displaystyle \L\:\frac{\frac{3}{2}(2x\,-\,3)\,+\,\frac{5}{2}}{x^2\,-\,3x\,+\,2} \;=\;\frac{3}{2}\cdot\frac{2x\,-\,3}{x^2\,-\,3x\,+\,2}\,+\,\frac{5}{2}\cdot\frac{1}{x^2\,-\,3x\,+\,2}\)

In the second denominator, complete the square:
. . \(\displaystyle \L x^2\,-\,3x\,+\,2 \;=\;x^2\,-\,3x\,+\,\frac{9}{4}\,+\,2\,-\,\frac{9}{4} \;=\;\left(x\,-\,\frac{3}{2}\right)^2\,-\,\frac{1}{4}\)

. . and we have: \(\displaystyle \L\:1\,+\,\frac{3}{2}\cdot\frac{2x\,-\,3}{x^2\,-\,3x\,+\,2}\,+\,\frac{5}{2}\cdot\frac{1}{(x\,-\,\frac{3}{2})^2 - \frac{1}{4}}\)


Integrate: \(\displaystyle \L\:\int 1\,dx \;+\;\frac{3}{2}\int\frac{2x\,-\,3}{x^2\,-\,3x\,+\,2} \;+\;\frac{5}{2}\int\frac{dx}{(x-\frac{3}{2})^2\,+\,\frac{1}{4}}\)

. . In the second integral, substitute: \(\displaystyle u\:=\:x^2\,-\,3x\,+\,2\)

. . In the third integral, substitute: \(\displaystyle v\:=\:x\,-\,\frac{3}{2}\)

\(\displaystyle \text{And we have: }\L\:\int1\,dx\;+\;\frac{3}{2}\underbrace{\int\frac{du}{u}}_{\ln}\,+\,\frac{5}{2}\underbrace{\int\frac{dv}{v^2\,+\,\frac{1}{4}}}_{\arctan}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Yes, I agree . . . it requires a lot of steps.
. . But the steps are from Algebra I only.

This technique is used when a quadratic denominator does not factor.

It is also a method guarenteed to impress/surprise/terrify your teacher.

You accidently replaced -1/4 by +1/4 in the denominator which leads to an arctan instead of a logarithm. This, I guess, is an illustration of the principle: "the shorter the derivation, the less likely it is to contain mistakes"