Evaluate the following integral x arctan x1 x22 dxSolutionm

Evaluate the following integral: x arctan x/(1 + x2)2 dx

make an integration by parts
u=arctan(x)
du=1/(1+x^2)
dv=x/(1+x^2)^2
v=-1/2*1/(1+x^2)

So you get

=-1/2*1/(1+x^2)*arctan(x)|₀-₀ -1/2*1/(1+x^2)^2 dx

The first term is 0

so we get

=1/2₀ 1/(1+x^2)^2 dx

now make substitution x=tan(w)

dx=sec^2(w)

you would get

=1/2cos^2(w) dw

=1/21/2(cos(2w)+1) dw

=1/4 (w+sin(2w)/2)

=1/4(arctan(x)+x/(x^2+1))

from 0 to infinity this is

=1/4*(/2)=/8