How Can You Solve the Integral of x*arctan(x)/(1+x^2)^2 from 0 to Infinity?

[1905]

this problem is in the improper integrals section but i just need help taking the integral. the bounds are from 0 to positive infinity.

xarctanx/(1+x^2)^2

i tried doing a trig sub x = tan θ, which gives θtanθ/(secθ)^2 I believe, which gives θsinθ/(cosθ)^3. I couldn't find anything to help me from here, so anything is appreciated. Perhaps I did one of the earlier steps wrong. Thanks in advance.

 

[Tide]

Try integration by parts noting that
\frac {x}{\left( 1 + x^2 \right)^2} = -\frac {1}{2} \frac {d}{dx} \frac {1}{1+x^2}

 

[diegojco]

integrate by parts

this integral is easy take a look:

you have int(x*arctanx/(((x^2)+1)^2)) well if you choose u=arctanx and dv=xdx/(((x^2)+1)^2) you will have that du=x/(1+(x^2)) and v=-1/2((x^2)+1) and since:

int(x*arctanx/(((x^2)+1)^2))=int(u*dv)=u*v-int(v*du) (integration by parts)

you can get the expresion:

int(x*arctanx/(((x^2)+1)^2))= (-arctanx-2)/(2((x^2)+1))