Question

How do you evaluate the integral `int arctanx/x^2`?

Answer 1

`-arctanx/x+lnabsx-1/2lnabs(x^2+1)+C`

Explanation:

`I=intarctanx/x^2dx`

Use integration by parts. Let:

`u=arctanx" "=>" "du=1/(x^2+1)dx`

`dv=1/x^2dx" "=>" "v=-1/x`

Then:

`I=uv-intvdu`

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

Performing partial fractions:

`1/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)`

`1=A(x^2+1)+(Bx+C)x`

`1=x^2(A+B)+x(C)+A`

Comparing coefficients,

`{(A+B=0),(C=0),(A=1):}`

So `B=-1` as well, giving the decomposition:

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

So:

`I=-arctanx/x+intdx/x-intx/(x^2+1)dx`

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

`I=-arctanx/x+lnabsx-1/2lnabs(x^2+1)+C`