Find $ \int \frac{dx}{(x^2 + a^2)^3} $

Question

I would like to find the anti-derivative

$$\displaystyle \int \dfrac{dx}{(x^2 + a^2)^3 }$$

My attempt:

By substitution: $ x = a \tan(\theta) \Rightarrow dx = a \sec^2(\theta) d\theta$

Then the integral becomes

$$\displaystyle \int \dfrac{ a \sec^2(\theta) d\theta }{a^6 \sec^6(\theta) } $$

And this reduces to

$$\dfrac{1}{a^5} \displaystyle \int \cos^4(\theta) d\theta $$

Now, $\cos^4(\theta) = ( \cos^2(\theta) )^2 = \dfrac{1}{4} (1 + \cos(2 \theta) )^2 = \dfrac{1}{4} ( 1 + 2 \cos(2 \theta) + \cos^2(2 \theta) ) $

So this reduces to

$$ \cos^4(\theta) = \dfrac{1}{4} + \dfrac{1}{2} \cos(2 \theta) + \dfrac{1}{8} (1 + \cos(4 \theta) ) $$

and these are integrable easily, but I want to relate their integrals to the original variable $x$.

So, now I have the anti-derivative as

$$ \dfrac{3}{8} \theta + \dfrac{1}{4} \sin(2 \theta) + \dfrac{1}{32} \sin(4 \theta) $$

Since $ x = a \tan(\theta) $, then $ \cos(\theta) = \dfrac{a}{\sqrt{x^2 + a^2}}$, and $\sin(\theta) = \dfrac{x}{\sqrt{x^2 + a^2}} $

Hence

$$\sin(2 \theta) = \dfrac{ 2 a x }{x^2 + a^2} $$

$$\cos(2 \theta) = \dfrac{a^2 - x^2}{x^2 + a^2} $$

And finally,

$$\sin(4 \theta) = \dfrac{4 a x (a^2 - x^2) }{ (x^2 + a^2)^2} $$

Hence, my integral becomes

$$ \displaystyle \int \dfrac{dx}{(x^2 + a^2)^3} = \dfrac{1}{a^5}\left( \dfrac{3}{8} \tan^{-1}\left(\dfrac{x}{a}\right) + \dfrac{1}{2} \dfrac{a x }{x^2 + a^2} + \dfrac{1}{8} \dfrac{ a x (a^2 - x^2) }{ (x^2 + a^2)^2 } \right) + C $$

where $C$ is an arbitrary constant of integration.

And my question is: Are my approach and derivation correct?

Answer 1

Alternatively, integrate by parts to avoid the trig substitution \begin{align} I_3=&\int \frac{1}{(x^2 + a^2)^3 }dx =\int \frac{1}{4a^2x^3}d\left(\frac{x^4}{ (x^2 + a^2)^2}\right) = \frac{x}{4a^2 (x^2 + a^2)^2}+\frac3{4a^2}I_2\\ I_2=&\int \frac{1}{(x^2 + a^2)^2}dx =\int \frac{1}{2a^2x}d\left(\frac{x^2}{ x^2 + a^2}\right) =\frac{x}{2a^2 (x^2 + a^2)}+\frac1{2a^2}I_1\\ I_1=&\int \frac{1}{x^2 + a^2}dx= \frac1a \tan^{-1}\frac xa \end{align} which leads to $$I_3 = \frac{x}{4a^2 (x^2 + a^2)^2}+ \frac{3x}{8a^4 (x^2 + a^2)}+ \frac{3}{8a^5}\tan^{-1}\frac xa +C $$

Answer 2

Furthermore, I am going to find the integral in general

$$ I_{n}=\int \frac{1}{\left(x^{2}+a^{2}\right)^{n}} d x $$

by a reduction formula by noting that

$$ \begin{aligned} \frac{d}{d x}\left[\frac{x}{\left(x^{2}+a^{2}\right)^{n}}\right] &=\frac{a^{2}+(1-2 n) x^{2}}{\left(x^{2}+a^{2}\right)^{n+1}} \\ &=\frac{1}{\left(x^{2}+a^{2}\right)^{n}}-2 n \cdot \frac{x^{2}+a^{2}-a^{2}}{\left(x^{2}+a^{2}\right)^{n+1}} \\ &=\frac{1}{\left(x^{2}+a^{2}\right)^{n}}-2 n\left[\frac{1}{\left(x^{2}+a^{2}\right)^{n}}-\frac{a^{2}}{\left(x^{2}+a^{2}\right)^{n+1}}\right] \end{aligned} $$ Integrating both sides w.r.t. $x$ yields $$ \boxed{I_{n+1}=\frac{x}{2 n a^{2}\left(x^{2}+a^{2}\right)^{n}}+\frac{2 n-1}{2 n a^{2}} I_{n}} $$ In particular, when $n=2$,

$$ \begin{aligned} I_{3} &=\frac{x}{4 a^{2}\left(x^{2}+a^{2}\right)^{2}}+\frac{3}{4 a^{2}} I_{2} \\ &=\frac{x}{4 a^{2}\left(x^{2}+a^{2}\right)^{2}}+\frac{3}{4 a^{2}}\left[\frac{x}{2 a^{2}\left(x^{2}+a^{2}\right)}+\frac{1}{2 a^{2}} I_{1}\right] \\ &=\boxed{\frac{x}{4 a^{2}\left(x^{2}+a^{2}\right)^{2}}+\frac{3 x}{8 a^{4}\left(x^{2}+a^{2}\right)}+\frac{3}{8 a^{5}} \tan ^{-1} \frac{x}{a}+C} \end{aligned} $$