How to solve this integration that I got from differential linear equation?

Question

I worked differential linear equation and at end of equation I got this integral.Can someone give me a hint to do this: $$\int \frac{1}{\left(u^2+1\right)^2}du$$

Answer 1

The trig substitution $u = \tan(x)$ is probably the best way to integrate this, but you can also integrate this by rewriting the integral as $$\int\frac{du}{(1+u^2)^2} = \int\Big(\frac{1}{1+u^2} + \frac{-u^2}{(1+u^2)^2}\Big)du$$ Edit: The second integral looked solvable without trig substitutions when I first typed it, but after looking at it again, a trig substitution looks unavoidable.

Answer 2

Substitute $u=\tan x$ and $du=\sec^2 x dx$, so $(u^2+1)^2=(\tan^2x+1)^2=\sec^4 x$ and $x=\arctan u$

$$=\int \cos^2 x dx$$

Now write $\cos^2 x$ as $\frac 1 2 \cos 2x+\frac 1 2$

$$=\int\left(\frac 1 2 \cos 2x+\frac 1 2\right)dx=\frac{\sin 2x}{4}+\frac x 2+C$$

Substitute back $x=\arctan u$

$$=\color{red}{\frac{u^2\arctan u+u+\arctan u}{2u^2+2}+C}$$

Answer 3

Substitue $u = \tan x$

Then it reduces to

$\int \cos^2 x \:\text{dx}$