# Homework Help: How do you tackle this integral

1. Sep 6, 2013

### Gauss M.D.

1. The problem statement, all variables and given/known data

Find F(x)

f(x) = 1/(1+x2)2)

2. Relevant equations

3. The attempt at a solution

It's actually a subproblem of another huge annoying surface integral. I tried u-sub but that landed me nowhere and partial integration led me nowhere aswell... Any pointers?

2. Sep 6, 2013

### Zondrina

Perfect candidate for a trig sub. Use $x = tan(θ)$ so that $dx = sec^2(θ)dθ$.

3. Sep 6, 2013

### ZeroPivot

well, 1/1+X^2 is the derivative of arctan i believe. i think there has to be a relation. u use integration by parts left u nowhere?

4. Sep 6, 2013

### Gauss M.D.

Zondrina: Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?

Last edited: Sep 6, 2013
5. Sep 6, 2013

### ZeroPivot

its a nasty integral u sure you are doing everything correctly? so the integral is 1/1+x^2 only?

6. Sep 6, 2013

### Gauss M.D.

No, I was referring to Zondrinas trig sub suggestion!

7. Sep 6, 2013

### Zondrina

I don't see any discontinuity in the integrand at all, so I don't think it would matter.

8. Sep 6, 2013

### Gauss M.D.

Right right, we're using tan and not some other trig function. My bad again. Thanks a ton :)

9. Sep 7, 2013

### verty

That is a very good question, I don't know the answer to it. I'm going to investigate this.

10. Sep 7, 2013

### verty

Sorry, the original question has no trouble because the range of tan is just R. Here is a more applicable problem:

$y = \int {dx \over (1 - x^2)^2}$

$x =? \; sin\theta$
$dx = cos\theta \; d\theta$

$y = \int sec^3\theta \; d\theta$
$= {1 \over 2} [sec\theta \; tan\theta + \int sec\theta \; d\theta] + C$
$= {1 \over 2} [sec\theta \; tan\theta + ln| sec\theta + tan\theta | ] + C$
$= {1\over 2}{x \over 1 - x^2} + {1\over 4} ln | {1+x\over 1-x}| + C$

On the other hand, one can make a translation:

$x = u-1$
$dx = du$

$y = \int {dx \over (1 - x^2)^2} = \int{ dx \over (1+x)^2 (1-x)^2} = \int {du \over u^2(2-u)^2}$

After some partial fraction magic:

$y = \int {1 \over 4(2-u)^2} + {1 \over 4u^2} + {1 \over 4u} + {1 \over 4(2-u)} du$
$= {1 \over 4} [ {1 \over 2-u} - {1\over u} + ln|u| - ln|2-u| ] + C$
$= {1 \over 4} [ {2(u-1) \over u(2-u)} + ln| {u \over 2-u} | ] + C$
$= {1 \over 2} {x \over 1 - x^2} + {1 \over 4} ln| {1+x \over 1-x} | + C$

I can't explain it but it's the same answer. Probably it has to do with the behaviour of sin(z) and cos(z) for complex z.