"Evaluate ∫1(x2+1)2dx" was answered by our experts

Adela Brown

Answered question

2022-01-05

Evaluate

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

Answer & Explanation

nghodlokl

Beginner2022-01-06Added 33 answers

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

Set

x = \tan (u)

and

d x = \sec^{2} (u) d u

. Then

{(x^{2} + 1)}^{2} = {(\tan^{2} (u) + 1)}^{2} = \sec^{4} (u)

and

u = \arctan (x)

= \int \cos^{2} (u) d u = \frac{1}{2} \int \cos (2 u) d u + \frac{1}{2} \int 1 d u

= \frac{u}{2} + \frac{1}{4} \sin (2 u) + C

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

Laura Worden

Beginner2022-01-07Added 45 answers

There is a faster way. Substitute

x = \tan (z) d x = \sec^{2} (z) d z

thus

{(x^{2} + 1)}^{2} = {(\tan^{2} (z) + 1)}^{2} = \sec^{4} (z) z = \arctan (x)

And remembering that

\frac{1}{\sec^{2}} = \cos^{2}

your integral is simply

\int \cos^{2} (z) d z

Which is trivial and left to you.
The tangent/secant substitution is a great technique which most of people ignore. Study it, and you will solve lots of awesome integrals!
Final result:

\frac{x^{2} \arctan (x) + x + \arctan (x)}{2 x^{2} + 2}

star233

Skilled2022-01-11Added 403 answers

Let us find $\frac{d [(a x + b) / (x^{2} + 1)^{n}]}{d x}$
$= \frac{- a x^{2} - 2 b x + a}{(x^{2} + 1)^{n + 1}} = \frac{- a (x^{2} + 1) - 2 b x + 2 a}{(x^{2} + 1)^{n + 1}} = - \frac{a}{(x^{2} + 1)^{n}} + \frac{2 a - 2 b x}{(x^{2} + 1)^{n + 1}}$
Integrating both sides,
$b = 0 \Rightarrow 2 \int \frac{d x}{(x^{2} + 1)^{n + 1}} = \int \frac{d x}{(x^{2} + 1)^{n}} + \frac{x}{(x^{2} + 1)^{n}} n = 1 \Rightarrow 2 \int \frac{d x}{(x^{2} + 1)^{2}} = \int \frac{d x}{x^{2} + 1} + \frac{x}{(x^{2} + 1)}$

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