`int_0^1 (dx)/(x^2 + 1)^2` Evaluate the integral

First find the indefinite integral:

`int (dx)/(1+x^2)^2=int (1+x^2-x^2)/(1+x^2)^2 dx = int (dx)/(1+x^2)-int (x^2 dx)/(1+x^2)^2.`

The first summand is `arctan(x),` for the second apply integration by parts:

`u=x,`  `dv=(x dx)/(1+x^2),` so `du=dx` and `v=-1/2 1/(1+x^2).`

So  `int (x^2 dx)/(1+x^2)^2=-1/2 x/(1+x^2)+1/2 int (dx)/(1+x^2) = -1/2 x/(1+x^2)+1/2arctan(x).`

Thus the indefinite integral is  `1/2(x/(1+x^2)+arctan(x))+C,`

and the indefinite integral is  `1/2(1/2+pi/4-0-0)=1/4+pi/8.`