Question

`int_0^oo arctanx/(x(x^2+a^2))dx` ?

Answer 1

`I=pi/(2a^2)ln(1+abs(a))`

Explanation:

We want to evaluate

`I=int_0^ooarctan(x)/(x(x^2+a^2))dx`

This can be solved by differentiation under the integral sign

Consider

`I(b)=int_0^ooarctan(bx)/(x(x^2+a^2))dx`

Notice `color(blue)(I(0)=0` and `color(blue)(I(1)=I`

Differentiate both sides w.r.t. b

`I'(b)=int_0^oo1/((x^2+a^2)(x^2b^2+1))dx`

By partial fractions

`I'(b)=1/(1-a^2b^2)int_0^oo1/(x^2+a^2)-b^2/(x^2b^2+1)dx`

We recognize these integrals as

`I'(b)=1/(1-a^2b^2)[arctan(x/a)/a-b arctan(bx)]_0^oo`

Assuming `color(blue)(a>0` and `color(blue)(b>=0`

(We could just as well assumed `color(blue)(a<0` )

`I'(b)=1/(1-a^2b^2)((pi/2)/a-bpi/2)`

By some algebraic manipulation

`I'(b)=pi/2*1/(1-a^2b^2)((1-ab)/a)`

`color(white)(I'(b))=pi/(2a)(1-ab)/(1-a^2b^2)`

`color(white)(I'(b))=pi/(2a)1/(1+ab)`

Integrate both sides w.r.t b

`I(b)=pi/(2a)int1/(1+ab)db`

`color(white)(I(b))=pi/(2a^2)ln(1+ab)+C`

Evaluate the constant of integration (Remember `color(blue)(I(0)=0`)

`0=pi/(2a^2)ln(1+a*0)+C=>C=0`

Thus

`I(b)=pi/(2a^2)ln(1+ab)`

Respecting negative values of `a` as well

`I(b)=pi/(2a^2)ln(1+abs(a)b)`

In your case `color(blue)(b=1`

`I(1)=I=pi/(2a^2)ln(1+abs(a))`