# [SOLVED] Partial fractions integral

• Jul 19th 2008, 02:39 PM
arbolis
[SOLVED] Partial fractions integral
I'm betting I have to solve it using partial fractions : $\int \frac{2x^3+3x^2+2x+4}{(x^2+1)^2}dx$. So I found $A$, $B$, $C$ and $D$ and I reached to that the solution must be equal to $\int \frac{2x+3}{x^2+1}+\frac{4}{(x^2+1)^2}dx=\ln |x^2+1|+4\int \frac{dx}{(x^2+1)^2}$. And I'm stuck here! This last integral looks quite simple I know, and is very similar to $\arctan (x)$, but I don't know how to solve it. I tried by integration by parts, u-substitution. Should I try for trigonometric substitution? If yes, please start me up making a good choice for the sub. I'm not used to trigo-sub... Thanks.
• Jul 19th 2008, 02:44 PM
Mathstud28
Quote:

Originally Posted by arbolis
I'm betting I have to solve it using partial fractions : $\int \frac{2x^3+3x^2+2x+4}{(x^2+1)^2}dx$. So I found $A$, $B$, $C$ and $D$ and I reached to that the solution must be equal to $\int \frac{2x+3}{x^2+1}+\frac{4}{(x^2+1)^2}dx=\ln |x^2+1|+4\int \frac{dx}{(x^2+1)^2}$. And I'm stuck here! This last integral looks quite simple I know, and is very similar to $\arctan (x)$, but I don't know how to solve it. I tried by integration by parts, u-substitution. Should I try for trigonometric substitution? If yes, please start me up making a good choice for the sub. I'm not used to trigo-sub... Thanks.

There is the first part and if you scroll down (you don't have to) there is a tad more but not the full solution, for I know that is not what you want.

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

Let $x=\tan(\theta)$

$dx=\sec^2(\theta)$

so we have

$\int\frac{sec^2(\theta)}{\sec^4(\theta)}~d\theta=\ int\cos^2(\theta)~d\theta$
• Jul 19th 2008, 02:54 PM
arbolis
Sorry mathstud28, I don't understand what you did. I've scrolled down what I wasn't obligated to, but this confuses me. We would have $\int \frac{dx}{x^2}$, which isn't true, isn't it?
And yes, you know me well, I prefer not to have the full answer if possible.
• Jul 19th 2008, 03:10 PM
arbolis
But I just realized I just showed a formula that fit exactly with this integral! Look at one of my earlier post...
So the integral equals to $\frac{x}{2x^2+2}+\frac{\arctan (x)}{2}+C$. Thus the final result follows.
• Jul 19th 2008, 03:20 PM
Mathstud28
Quote:

Originally Posted by arbolis
But I just realized I just showed a formula that fit exactly with this integral! Look at one of my earlier post...
So the integral equals to $\frac{x}{2x^2+2}+\frac{\arctan (x)}{2}+C$. Thus the final result follows.

That is the correct answer, here is the trig sub justification.

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

If we let

$x=\tan(\theta)$

$\Rightarrow{dx=\sec^2(\theta)}d\theta$

So we would have

$\int\frac{sec^2(\theta)}{(\tan^2(\theta)+1)^2}~d\t heta$

$=\int\frac{\sec^2(\theta)}{(\sec^2(\theta))^2}~d\t heta$

$=\int\frac{d\theta}{\sec^2(\theta)}$

$=\int\cos^2(\theta)~d\theta$

$=\frac{1}{2}\int\bigg[1+\cos(2\theta)\bigg]~d\theta$

$=\frac{1}{2}\bigg[\theta+\frac{1}{2}\sin(2\theta)\bigg]$

$=\frac{1}{2}\bigg[\theta+\sin(\theta)\cos(\theta)\bigg]$

$\underbrace{=}_{\text{backsub}}\frac{1}{2}\bigg[\arctan(x)+\sin\left(\arctan(x)\right)\cos\left(\a rctan(x)\right)\bigg]$

$=\frac{1}{2}\bigg[\arctan(x)+\frac{x}{\sqrt{1+x^2}}\cdot\frac{1}{\sq rt{1+x^2}}\bigg]$

$=\frac{1}{2}\bigg[\arctan(x)+\frac{x}{x^2+1}\bigg]$

$=\frac{\arctan(x)}{2}+\frac{x}{2(x^2+1)}$

That ought to be detailed enough for you :D.
• Jul 19th 2008, 03:28 PM
arbolis
Quote:

That ought to be detailed enough for you :D.
Yes! Let me analyze this now.
Note that in your first post you said and in your last : . (Hi)
• Jul 19th 2008, 03:31 PM
Mathstud28
Quote:

Originally Posted by arbolis
Yes! Let me analyze this now.
Note that in your first you said and in your last : . (Hi)

Oops. (Smirk)..now I see why you might have been confused (Giggle). Desole.