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

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?

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?

Alexx1
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?

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

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

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.

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Alexx1
Thank you both!

Dunebug7
Thank you both!

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.

fourier jr
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?

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:

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