Integral trigonometric with arctan

Question

The integral in question to solve is: $$ \int{ \frac{\arctan^3(x)}{1+x^2}dx } $$

I try solve this for two methods, first time with only u.du substituition and du/dx dont is replaceble in integral. The second try I use u.dv method and I get an recursive u.du integral before solve the first u.dv.

Answer 1

use the substitution:

$u=arctan (x)$

so that

$$du=\frac{1}{1+x^2}dx$$ and the integral becomes:

$$ \int u^3 du $$

solve with propertie $\int {u^ndu}=\frac{u^n+1}{n+1}$: $$ \int{u^3du} = \frac{u^4}{4}+C $$

back to original variable $x$:

$$ \frac{u^4}{4}+C = \frac{arctan⁴(x)}{4}+C $$

Answer 2

$$I=\int \dfrac {\arctan^3(x)}{1+x^2}dx$$ is of the form $$J=\int f^3(x)f'(x)dx$$ $$J=\frac 1 4 \int (f^4(x))'dx$$ $$J=\frac 14 f^4(x)+C$$ $$\implies I=\frac14 \arctan^4(x)+C$$

Answer 3

I just wanted to point out that the other method attempted also works. Let $u=\left(\tan^{-1}x\right)^3$, $dv=\frac{dx}{1+x^2}$. Then $du=3\left(\tan^{-1}x\right)^2\frac{dx}{1+x^2}$, $v=\tan^{-1}x$ so $$\int\frac{\left(\tan^{-1}x\right)^3}{1+x^2}dx=\int u\,dv=uv-\int v\,du=\left(\tan^{-1}x\right)^3\left(\tan^{-1}x\right)-\int\left(\tan^{-1}x\right)\times3\frac{\left(\tan^{-1}x\right)^2}{1+x^2}dx$$ On addition of $3\int\frac{\left(\tan^{-1}x\right)^3}{1+x^2}dx$ to both sides we get $$4\int\frac{\left(\tan^{-1}x\right)^3}{1+x^2}dx=\left(\tan^{-1}x\right)^4+C$$