Question

How do you integrate `int ln(1+x^2)` by parts from `[0,1]`?

Answer 1

`int_0^1 ln(1+x^2) dx =ln2 -2 +pi/2`

Explanation:

We can take `x` as finite part:

`int_0^1 ln(1+x^2) dx = [xln(1+x^2)]_0^1 - int_0^1 x d(ln(1+x^2))`

`(1) int_0^1 ln(1+x^2) dx =ln2 - 2int_0^1 (x^2dx)/(1+x^2)`

Now solving the resulting integral:

`int_0^1 (x^2dx)/(1+x^2) = int_0^1 (x^2+1-1)/(1+x^2)dx`

`int_0^1 (x^2dx)/(1+x^2) = int_0^1 (1-1/(1+x^2))dx`

`int_0^1 (x^2dx)/(1+x^2) = int_0^1 dx - int_0^1 dx/(1+x^2)`

`int_0^1 (x^2dx)/(1+x^2) = [x]_0^1-[arctanx]_0^1`

`int_0^1 (x^2dx)/(1+x^2) = 1-pi/4`

Substitute in (1):

`int_0^1 ln(1+x^2) dx =ln2 -2 +pi/2`