How to integrate $\int_0^\infty\frac{dx}{(1+x^2)^2}$?

Question

I need to integrate this to finish an old STEP problem I'm doing, but I'm stuck here, at the very end:

$$\int_0^\infty \frac{dx}{(1+x^2)^2}$$

The result should be $\pi\over 4$ . I don't know how to approach this. *Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).

Also, Wolfram tells me:

$$\int \frac{dx}{(1+x^2)^2} = \frac{1}{2}\left(\frac{x}{x^2+1}+\tan^{-1}x\right)+c$$

but I don't see how one can derive this without knowing the result beforehand.

Please, help me!

Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).

EDIT: If you're interested, the problem in question is: STEP II - problem 4 (year 2014).

use $\frac 1{(1+x^2)^2 }=\frac 1{1+x^2} - \frac{x^2}{(1+x^2)^2} $ and do an integration by parts $\int \frac{x^2}{(1+x^2)^2} , dx$ — abel, Jun 02 '16 at 18:04
I tried $x = \sinh(z)$ since $1+x^2 = \cosh^2(x)$ but I got stuck at $$\int \frac{1}{\cosh^3(z)} , {\rm d}z$$ — John Alexiou, Jun 02 '16 at 18:12
@JohnAlexiou put $t=\tanh z$, then what remains is $\int\sqrt{1-t^2}\mathrm dt$. — Integreek, Dec 07 '24 at 14:32
If you want to look up math formulae, use the Approach0.xyz search engine. — Integreek, Dec 07 '24 at 14:33

Jack D'Aurizio · Answer 1 · 2024-12-30T23:13:45.743

9

For any $\alpha>0$, let: $$ I(\alpha)= \int_{0}^{+\infty}\frac{dx}{\alpha^2+x^2} = \frac{\pi}{2\alpha}.$$ By differentiating both sides with respect to $\alpha$ we get: $$ \int_{0}^{+\infty}\frac{2\alpha}{(\alpha^2+x^2)^2}\,dx = \frac{\pi}{2\alpha^2} $$ and by evaluating at $\alpha=1$: $$ \int_{0}^{+\infty}\frac{dx}{(x^2+1)^2} = \color{red}{\frac{\pi}{4}}.$$

edited Dec 30 '24 at 23:13

answered Jun 02 '16 at 18:03

Jack D'Aurizio

361,689

How do you get these expressions? – John Alexiou Jun 02 '16 at 18:07
3

@ja72: What's obscure? I think my answer is pretty straightforward to follow. The technique is known as differentiation under the integral sign, or "Feynman's trick". – Jack D'Aurizio Jun 02 '16 at 18:08
I'm sure that's common knowledge, but how (a very general idea would do) do you prove that differentiating an integral is the same as differentiating the thing that's integrated? – Kuba Jun 02 '16 at 18:09
1

@ja72 Leibniz integral rule. – M. Vinay Jun 02 '16 at 18:10
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@M.Vinay: thanks, I was just writing the same thing :) – Jack D'Aurizio Jun 02 '16 at 18:12
I as talking mostly of the first expression. How do you get $\int \ldots = \frac{\pi}{2 \alpha}$ – John Alexiou Jun 02 '16 at 18:15
@ja72: Through the substitution $x=\alpha z$ it boils down to $$\int_{0}^{+\infty}\frac{dz}{z^2+1}=\frac{\pi}{2}.$$ – Jack D'Aurizio Jun 02 '16 at 18:16
@JackD'Aurizio. Great, so maybe include this to make your answer more complete. Most of us don't have integration tables memorized. – John Alexiou Jun 02 '16 at 18:18
1

@ja72: I assure you that my memorization skills are close to being awful (that is the main reason beyond my appraise for modern technology, that allow us to store tons of useful data in portable devices), but when it comes to mathematical practice, many things become natural. Practice makes perfect. – Jack D'Aurizio Jun 02 '16 at 18:20
The method you added has already been mentioned in two answers. I don't see the point of mentioning it again? – Integreek Dec 15 '24 at 17:48

score 6 · Answer 2 · edited Dec 13 '24 at 10:46

6

Use $u=\tan x$ $$(1+x^2)^2=(1+\tan^2u)^2=(\sec^2u)^2=\sec^4u$$ $$\mathrm dx=\sec^2u\mathrm du$$ The integral becomes $$\int{\frac{1}{\sec^2u}\mathrm du=\int{\cos^2u\mathrm du}}$$

edited Dec 13 '24 at 10:46

Integreek

8,530

answered Jun 02 '16 at 18:12

Brittany

176

3

functions need to be upright characters. Use \sin instead of sin. Also the differential needs to be upright. Use {\rm d} instead of d. – John Alexiou Jun 02 '16 at 18:14

score 4 · Accepted Answer · answered Jun 02 '16 at 18:03

4

Use $x = \tan t$. Then $(1 + x^2)^2 = \sec^4 t$, and $\dfrac{dx}{dt} = \sec^2 t$, so the integral becomes \begin{align*} \int_0^{\pi/2} \cos^2 t \,\mathrm dt = \dfrac 1 2 \times \dfrac {\pi} 2 = \dfrac{\pi}{4}. \end{align*}

answered Jun 02 '16 at 18:03

M. Vinay

9,214

2

Nice. Thanks. It's been a while and I forgot about trig substitions :( – Kuba Jun 02 '16 at 18:08
1

$\displaystyle\int_0^{\pi/2}\cos^2x~\mathrm dx=\int_0^{\pi/2}\sin^2x~\mathrm dx$ can be used to evaluate the integral in a single step. – Integreek May 05 '25 at 03:43
@Integreek Yes, that can be used instead of the reduction formula. – M. Vinay May 06 '25 at 06:19

score 4 · Answer 4 · answered Jun 02 '16 at 19:43

4

let $I_n=\int_{0}^{\infty}\frac{1}{(1+x^2)^n}dx$ then $$I_{n+1}=\frac{2n-1}{2n}I_n$$ we know $I_1=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}$, so $$I_2=\frac{2(1)-1}{2(1)}I_1=\frac{\pi}{4}$$

answered Jun 02 '16 at 19:43

Behrouz Maleki

11,342

(+1) I see in you a future star of MSE. – Jack D'Aurizio Jun 02 '16 at 20:54

score 3 · Answer 5 · answered Dec 07 '24 at 14:26

Using the substitution $x\to\frac1{x}$ simplifies the integrand beautifully(I'm astonished that nobody thought of this before):

$$\mathcal I=\int_0^\infty\frac{\mathrm dx}{(x^2+1)^2}=\int_0^\infty\frac{x^2}{(x^2+1)^2}\mathrm dx$$

Hence,

$$\mathcal I=\frac12\int_0^\infty\frac{\mathrm dx}{x^2+1}=\frac\pi4$$

score 2 · Answer 6 · answered Jun 02 '16 at 18:03

2

Add and subtract $x^2$ from the numerator, you get $\arctan$ on one hand and an $\int \frac{x^2}{(1 + x^2)^2}dx$, which you can do by parts ($u = x$ and $dv = \frac{x}{(1+x^2)^2}dx$)

answered Jun 02 '16 at 18:03

score 2 · Answer 7 · answered Jun 04 '16 at 04:35

Here's a pretty solution. Notice that $$ I=\int_{0}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{\partial\Omega}\frac{1}{(1-iz)^2(1+iz)^2}dz$$ where $\Omega$ is the top part of the complex plane. Since $f(z)$ decays faster than $\frac{1}{z^2}$, $$I=\pi i \lim_{z \to i}\frac{d}{dz}\frac{(z-i)^2}{(1-iz)^2(1+iz)^2}=\frac{\pi}{4}$$.

score 2 · Answer 8 · answered Sep 07 '22 at 02:01

$$ \begin{aligned} \int \frac{d x}{\left(1+x^2\right)^2} &=-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{1+x^2}\right) \\ &=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(-\frac{1}{x^2}\right) \frac{1}{1+x^2} d x \\ &=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(\frac{1}{1+x^2}-\frac{1}{x^2}\right) d x \\ &=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \tan ^{-1} x+\frac{1}{2 x}+C \\ &=\frac{1}{2}\left(\frac{x}{1+x^2}+\tan ^{-1} x\right)+C \end{aligned} $$

For definite integral, we have $$ \begin{aligned} \int_0^{\infty} \frac{d x}{\left(1+x^2\right)^2} &=\frac{1}{2}\left[\frac{x}{1+x^2}+\tan ^{-1} x\right]_0^{\infty} =\frac{\pi}{4} \end{aligned} $$

score 1 · Answer 9 · answered Jun 02 '16 at 18:16

$$\int_0^\infty\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=\lim_{n\to\infty}\int_0^n\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=$$

Substitute $x=\tan(u)$ and $\text{d}x=\sec^2(u)\space\text{d}u$.

So $\left(1+x^2\right)^2=\left(1+\tan^2(u)\right)^2=\sec^4(u)$ and $u=\arctan(x)$.

This gives a new lower bound $u=\arctan(0)=0$ and upper bound $u=\arctan(n)$:

$$\lim_{n\to\infty}\int_0^{\arctan(n)}\cos^2(u)\space\text{d}u=$$

Use:

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

$$\frac{1}{2}\lim_{n\to\infty}\left[\int_0^{\arctan(n)}1\space\text{d}u+\int_0^{\arctan(n)}\cos(2u)\space\text{d}u\right]=$$

Substitute $s=2u$ and $\text{d}s=2\space\text{d}u$.

This gives a new lower bound $s=2\cdot0=0$ and upper bound $s=2\arctan(n)$:

$$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\int_0^{2\arctan(n)}\cos(s)\space\text{d}s\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\left[\sin(s)\right]_0^{2\arctan(n)}\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\left(\arctan(n)-0\right)+\frac{\sin(2\arctan(n))-\sin(0)}{2}\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\arctan(n)+\frac{\sin(2\arctan(n))}{2}\right]=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}$$

score 1 · Answer 10 · answered Mar 17 '25 at 09:32

Here's an elegant solution using Glasser's master theorem:

$$\begin{align}\int_0^\infty\frac{\mathrm dx}{(x^2+1)^2}&\overset{x\to\frac1x}=\int_0^\infty\frac{\mathrm dx}{\left(x-\frac1x\right)^2+4}\\&=\int_0^\infty\frac{\mathrm dx}{x^2+4}\\&=\frac\pi4\end{align}$$

How to integrate $\int_0^\infty\frac{dx}{(1+x^2)^2}$?

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