The correct options are
A ϕ(x)=1+x22
D Ψ(x)=−1+x22
Given, ∫xlog(1+x2)dx=ϕ(x).log(1+x2)+Ψ(x)+c
Consider, ∫xlog(1+x2)dx
=log(1+x2)x22−∫(2x1+x2)x22dx
=log(1+x2)x22−∫x31+x2dx
=x22log(1+x2)+∫(−x+xx2+1)dx
=x22log(1+x2)−x22+∫xx2+1dx
Substitute x2+1=t
⇒xdx=dt2
=x22log(1+x2)−x22+12∫1tdt
=x22log(1+x2)−x22+12log(1+x2)+C
=x22log(1+x2)−x22+12log(1+x2)+C
=x2+12log(1+x2)−x2+12+C′
On comparing with (1), we get
ϕ(x)=x2+12,ψ(x)=−x2+12