Question

How do you evaluate the integral `int ln(1+x^2)`?

Answer 1

`intln(1+x^2)dx=xln(1+x^2)-2x+2arctan(x)+C`

Explanation:

`I=intln(1+x^2)dx`

Integration by parts takes the form `intudv=uv-intvdu`. For the given integral, let:

`u=ln(1+x^2)`

`color(white)(u=ln(1+x^2))du=(2x)/(1+x^2)dx`

`dv=dx`

`color(white)(dv=dx)v=x`

Then:

`I=uv-intvdu`

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

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

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

`I=xln(1+x^2)-2intdx+2intdx/(1+x^2)`

`I=xln(1+x^2)-2x+2arctan(x)+C`