# Integral : (x^2)/((x^2+1)^2)

1. Dec 25, 2009

### Alexx1

The answer of the integral of (x^2)/((x^2+1)^2) is (1/2)(arctan(x)-(x/x^2+1))

In class, we've seen the steps to solve this integral, but I don't understand certain steps..
Can someone explain me how to solve this integral, step by step?

2. Dec 25, 2009

### uart

If you can post the steps and point out those that you didn't understand then I'm sure someone can help you.

BTW. The easiest way to do that one is "integration by parts". Have you learnt this technique yet?

3. Dec 25, 2009

### Alexx1

Sure, no problem, here are the steps:

Integral((x^2)/((x^2+1)^2)dx)
= (1/2)*Integral(x d(1/(x^2+1))
= (1/2)*(x/(x^2+1))-(1/2)*Integral(1/(x^2+1)dx)
= (1/2)*(x/(x^2+1))-(arctan(x))/2

(The answer I said earlier was wrong, this is the correct answer:(1/2)*(x/(x^2+1))-(arctan(x))/2)

Thank you

4. Dec 25, 2009

### Bohrok

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

Using ∫u v' = uv - ∫v u',
let u = x and v' = x/(x2 + 1)2

5. Dec 25, 2009

### Dunebug7

It's basically separating it into parts ie.

$\int \frac{x^2}{(x^2+1)^2}\rightarrow \int \frac{x}{1}.\frac{x}{(x^2+1)^2}\equiv x(x. \sin(\arctan(x)))$

as

$\frac{x}{1}=\frac{1}{2}x^2$

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

By the trig identity.

$\int\frac{x^2}{(x^2+1)^2}=-\frac{1}{2}.\frac{x}{(x^2+1)}+\frac{1}{2}\arctan(x)+C$

Don't forget the constant of integration, it's a silly way to loose marks.

Last edited: Dec 25, 2009
6. Dec 25, 2009

### Alexx1

Thank you both!

7. Dec 25, 2009

### Dunebug7

np Bhorok's answer is more elegant and easier, but I thought you might need a long winded explanation and there's often more than one way to swing a cat I guess. Hope it helped.

8. Dec 25, 2009

### fourier jr

since you have the answer, take its derivative & work backwards. that's how to figure it out. just don't show anyone your rough work :tongue2:

9. Dec 25, 2009

### uart

No the original answer was correct, you dropped a minus sign in the first line of this derivation.