# Integration Partial Fractions

1. Sep 6, 2012

### 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 dont 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?

2. Sep 6, 2012

### Muphrid

Find the common denominator.

3. Sep 6, 2012

### phyzguy

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

4. Sep 6, 2012

### Polymath89

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