int (x arctan(x))/(1+x^2)^2 dx∫xarctan(x)(1+x2)2dx
= int arctan(x) (x )/(1+x^2)^2 dx=∫arctan(x)x(1+x2)2dx
= int arctan(x) ( (- 1/2 )/(1+x^2) )^' dx
= -1/2 int arctan(x) ( (1)/(1+x^2) )^' dx
which by IBP
= -1/2 ( arctan(x) (1)/(1+x^2) - int ( arctan(x) )^' (1)/(1+x^2) dx )
= -1/2 ( arctan(x) (1)/(1+x^2) - color(blue)( int 1/(1+x^2)^2 dx) ) qquad square
for the blue bit we say that x = tan xi, dx = sec^2 xi \ d xi
So we have
int 1/(1+tan^2 xi)^2 sec^2 xi \ d xi = int cos^2 xi \ d xi
for this we use the double angle formula: cos 2 gamma = 2 cos^2 gamma - 1 giving us:
1/2 int cos 2 xi + 1 \ d xi
= 1/2 ( 1/2 sin 2 xi + xi ) + C
= 1/2 ( sin xi cos xi + xi ) + C qquad triangle
and if tan xi = x, then sin xi = x/sqrt(x^2 +1) and cos xi = 1/sqrt(x^2 +1), just draw the right-angled triangle to see.
so triangle becomes
= 1/2 ( x/(x^2 +1) + arctan x ) + C qquad triangle
So when we pop it back into square we get
= -1/2 ( arctan(x) (1)/(1+x^2) - (1/2 ( x/(x^2 +1) + arctan x ) + C )
= -1/2 arctan(x) (1)/(x^2+1) + 1/4 x/(x^2 +1) + 1/4 arctan x + C
Tidying up
=1/4 1/(x^2+1) ( -2 arctan(x) + x + (x^2 +1) arctan x ) + C
=1/4 1/(x^2+1) ( x + (x^2 -1) arctan x ) + C
= ( x + (x^2 -1) arctan x )/(4(x^2+1)) + C
