# $\int\frac{n}{(n^{2}+1)^{2}}$= itself w/ Partial Fractions

1. Nov 8, 2013

### Lebombo

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

Why when I try to evaluate this with Partial Fractions, why do I end up with the original function?

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

$\frac{n}{(n^{2}+1)(n^{2}+1)}$

$\frac{Ax+B}{n^{2}+1} + \frac{Cx+D}{(n^{2}+1)^{2}}$

1n = (An+B)(n^{2}+1) + Cx + D

$0n^{3}+ 0n^{2} + 1n + 0n^{0} = n^{3}(A) + n^{2}(B) + n(A+C) + n^{0}(B+D)$

A=0 B=0 C=1 D=0

$\frac{0n+0}{n^{2}+1} + \frac{1n+0}{(n^{2}+1)^{2}}$

= 0 + $\frac{n}{(n^{2}+1)^{2}}$

2. Nov 8, 2013

3. Nov 8, 2013

### Lebombo

That website looks all scrambled up. Nothing seems to be aligned. Which part pertains to why this particular function ends up to simply be itself?

By the way, when I do u-sub on this function I get 1/4 and when I do partial fractions I get -1/4. Is that a sign mistake?

Also, the basis for this question was actually in regards to the series of the function in question.

I was trying to do partial fractions to create a telescoping function to evaluate the series. But partial fractions doesn't separate this function into telescoping series.

So my follow up questions is: Why does this function not separate into a telescoping form if Partial fractions can be applied to it?

4. Nov 8, 2013

### Staff: Mentor

Lebombo,
What you're saying doesn't make a lot of sense. As you've seen, partial fractions leads you right back to the exact same integrand, so I don't see how you were able to carry out the integration that way. Substitution is the way to go here.

The integral looks like it would be amenable to a trig substitution, but that seems like a lot of work when an ordinary substitution would do the trick.

5. Nov 8, 2013

### UltrafastPED

The site works fine with Chrome on my laptop. Check your browser!

The web site page goes over the method in detail, with a few examples. A review of the method may help you to find any problems with your technique.

Wolfram Alpha says: $n^2/(n^2+1) = 1-1/(n^2+1)$

6. Nov 9, 2013

### SteamKing

Staff Emeritus
IDK why you want to use partial fractions anyway. You've plainly got a case for a u substitution, no partial fractions or trig substitution required.

7. Nov 9, 2013

### Saitama

I don't see the need of a trig substitution. There is a very obvious substitution you can use as SteamKing has said.

8. Nov 9, 2013

### Lebombo

I just [thought I] saw an opportunity to practice a partial fraction from the past chapter while working on series.

Thanks all.