How to calculate $\int \frac{dx}{(a^2 + x^2)^2}$?

Question

I'm trying to use a trig substitution but I'm stuck. Here's what I did so far:

$$\int \frac{dx}{(a^2 + x^2)^2}$$

Let $x = a\sin \theta, dx = a\cos \theta d\theta$

$$\int \frac{a cos\theta d\theta}{(a^2 + a^2 sin^2 \theta)^2} = \int \frac{a\cos \theta d\theta}{(a^2(1+sin^2\theta))^2} $$

$$\int \frac{a\cos \theta d\theta}{(a^2 \cos^2\theta)^2} =\int \frac{d\theta}{a^3cos^3\theta} $$

I don't know what to do anymore

Answer 1

Substituting $x=a\tan(u)$ gives $$ \begin{align} \int\frac{\mathrm{d}x}{\left(a^2+x^2\right)^2} &=\frac1{a^3}\int\frac{\sec^2(u)\,\mathrm{d}u}{\sec^4(u)}\\ &=\frac1{a^3}\int\cos^2(u)\,\mathrm{d}u\\ &=\frac1{2a^3}\int(1+\cos(2u))\,\mathrm{d}u\\ &=\frac1{2a^3}(u+\sin(u)\cos(u))+C\\ &=\frac1{2a^3}\left(\tan^{-1}\left(\frac xa\right)+\frac{ax}{a^2+x^2}\right)+C\\ \end{align} $$

Answer 2

You can use that $$ \int\frac{\text{d}x}{a^2+x^2}=\frac{1}{a}\text{arctan}\left(\frac{x}{a}\right)+K $$ And then integrating by parts gives you

$$ \int\frac{\text{d}x}{a^2+x^2}=\left[\frac{x}{a^2+x^2}\right]+\int \frac{2x^2}{\left(a^2+x^2\right)^2}\text{d}x$$ $$=\left[\frac{x}{a^2+x^2}\right]+2\int \frac{\text{d}x}{a^2+x^2}-2a^2\int \frac{\text{d}x}{\left(a^2+x^2\right)^2} $$ Then $$ 2a^3\int \frac{\text{d}x}{\left(a^2+x^2\right)^2}=\left[\frac{ax}{a^2+x^2}\right]+\text{arctan}\left(\frac{x}{a}\right) $$ You finally obtain

$$ \int \frac{\text{d}x}{\left(a^2+x^2\right)^2}=\frac{1}{2a^3}\left(\frac{ax}{a^2+x^2}+\text{arctan}\left(\frac{x}{a}\right)\right)+C $$

Answer 3

observe that $$a^2+a^2\tan^2(t)=a^2(1+\tan^2(t))=\frac{a^2}{\cot^2(t)}$$

Answer 4

This could be done in a "general" way.

Denoting $$ I_k=\int \frac{dx}{(x^2+a^2)^k}, $$ one has (for $k>1$) $$ \begin{align} I_k&=\frac{1}{a^2}\int\frac{(x^2+a^2)-x^2}{(x^2+a^2)^k}\ dx\\ &=\frac{1}{a^2} I_{k-1}-\frac{1}{a^2}\int\frac{x^2}{(x^2+a^2)^k}\ dx\\ &=\frac{1}{a^2} I_{k-1}+\frac{1}{2a^2(k-1)}\int x\ d\left(\frac{1}{(x^2+a^2)^{k-1}}\right)\\ &=\frac{1}{a^2} I_{k-1}+\frac{1}{2a^2(k-1)}\left[\frac{x}{(x^2+a^2)^{k-1}}-I_{k-1}\right], \end{align} $$ which gives $$ I_k=\frac{x}{2a^2(k-1)(x^2+a^2)^{k-1}}+\frac{2k-3}{2a^2(k-1)}I_{k-1}. $$ Now you can get $I_2$ by observing that $$ I_1=\frac{1}{a}\arctan\frac{x}{a}+C. $$

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^{Notes: The recursive relation above allows one to calculate $I_k$ for any positive integer $k$ in principle, which can be used to calculate any indefinite integral of rational functions.}

Answer 5

$$\int\dfrac{dx}{a^2+x^2}=\dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right)+C$$ then $$\dfrac{d}{da}\int\dfrac{dx}{a^2+x^2}=\dfrac{d}{da}\left(\dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right)+C\right)$$ and $$\int\dfrac{-2adx}{(a^2+x^2)^2}=\dfrac{1}{2a^3}\arctan\left(\dfrac{x}{a}\right)-\dfrac{x}{2a^2(a^2+x^2)}$$

Answer 6

For any $A>0$ we have $$ \int\frac{dx}{A+x^2}=C+\frac{1}{\sqrt{A}}\arctan\frac{x}{\sqrt{A}}$$ hence by differentiating both sides with respect to $A$ we get $$ \int \frac{dx}{(x^2+A)^2} = D+\frac{x}{2A(x^2+A)}+\frac{1}{2A\sqrt{A}}\arctan\frac{x}{\sqrt{A}} $$ and now we just have to evaluate both sides at $A=a^2$.