We have an answer to "How to calculate ∫x2(x2+1)2dx"

Julia White

Answered question

2022-01-03

How to calculate

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

Answer & Explanation

Terry Ray

Beginner2022-01-04Added 50 answers

First, you used parts: with

u = x

and

d v = \frac{x}{{(x^{2} + 1)}^{2}} d x

, you got

\int \frac{x^{2}}{{(x^{2} + 1)}^{2}} d x = x v - \int v d x

(1)
From here, I suggested the substitution

t = x^{2}

in order to integrate dv and get v. Doing this, we see that

d t = 2 x d x

, so

x d x = \frac{1}{2} d t

. Thus

v = \int d v = \int \frac{x}{{(x^{2} + 1)}^{2}} d x = \frac{1}{2} \int \frac{d t}{{(t + 1)}^{2}}

This latter integral is easy: if you can't guess it, let

s = t + 1

. In any case, we get

v = - \frac{1}{2 (t + 1)} = - \frac{1}{2 (x^{2} + 1)}

Going back to (1),

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

and you should recognize the final integral as

\arctan x

. Putting it all together, the answer is

\frac{1}{2} \arctan x - \frac{x}{2 (x^{2} + 1)} + C

Daniel Cormack

Beginner2022-01-05Added 34 answers

HINT:
Use trigonometric substitution

x = \tan θ

\int \frac{x^{2}}{{(x^{2} + 1)}^{2}} d x = \frac{1}{2} \int (1 - \cos 2 θ) d θ

\sin 2 θ = \frac{2 \tan θ}{1 + \tan^{2} θ}

Vasquez

Expert2022-01-09Added 669 answers

$\begin{matrix} \int \frac{x^{2}}{(x^{2} + 1)^{2}} d x \\ = x \int \frac{x}{(x^{2} + 1)^{2}} d x - \int [\frac{d x}{d x} \cdot \int \frac{x}{(x^{2} + 1)^{2}} d x] d x \\ = x \cdot \frac{- 1}{1 + x^{2}} + \int \frac{d x}{1 + x^{2}} = . . . \end{matrix}$

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