What is the integration step used for quadratic factors in the denominator?

[Polymath89]

Im reading Lang's first course in calculus and can't understand one step that he does when trying to integrate quotients with quadratic factors in the denominator. He's trying to find the integral of $$\int{\frac{1}{(x^2+1)^n}dx}$$

but he's first starting with the case where n=1

Then while using integration by parts he gets this integral for $\int{vdu}$ $$2 \int{\frac{x^2}{(x^2+1)^2}dx}$$

then he writes x^2=x^2+1-1 and gets $$\int{\frac{1}{x^2+1}dx}-\int{\frac{1}{(x^2+1)^2}dx}$$

now I don't understand why he gets those two integrals as a result of writing x^2 as x^2+1-1, can anybody please help me out here?

 

[Muphrid]

Find the common denominator.

 

[phyzguy]

$$\frac{x^2}{(x^2+1)^2} = \frac{x^2+1-1}{(x^2+1)^2} = \frac{(x^2+1)-1}{(x^2+1)^2} = \frac{x^2+1}{(x^2+1)^2} - \frac{1}{(x^2+1)^2} = \frac{1}{x^2+1} - \frac{1}{(x^2+1)^2}$$

 

[Polymath89]

sorry didn't see that^^ thanks a lot guys.