How do I solve this trigonometric substitution integral?

Question

I know for a fact that it's a trig substitution where u=tanØ (let's pretend Ø is theta) and u=x and a=1.

For some reason, I keep going in circles.

This is the integral that I need to solve

Answer 1

$$\int\dfrac{2x^2}{(x^2+1)^2}dx=\int\dfrac{2(x^2+1-1)}{(x^2+1)^2}dx$$

$$=2\int\dfrac1{x^2+1}dx-\int\dfrac{2dx}{(x^2+1)^2}$$

For the second integral choose $x=\tan y\implies dx=\sec^2y\ dy$

$$\int\dfrac{2dx}{(x^2+1)^2}=\cdots=\int(1+\cos2y)dy=?$$

$$\sin2y=\dfrac{2\tan y}{1+\tan^2y}=?$$

Answer 2

Substitute $x=\tan (\theta)$ so $dx=\sec^2 (\theta)d\theta$. Use the identity $\tan^2 (\theta)+1=\sec^2 (\theta)=\frac {1}{\cos^2 (\theta)} $.