Partial fractions of x2+2x(x2+1)2, will be of type
x2+2x(x2+1)2=Ax+Bx2+1+Cx+D(x2+1)2
= (Ax+B)(x2+1)+Cx+D(x2+1)2
= Ax3+Bx2+Ax+B+Cx+D(x2+1)2
= Ax3+Bx2+(A+C)x+(B+D)(x2+1)2
Comparing the terms, we get A=0, B=1, A+C=2 and B+D=0
i.e. A=0, B=1, C=2 and D=−1
Hence, x2+2x(x2+1)2=1x2+1+2x−1(x2+1)2 and
∫x2+2x(x2+1)2dx=∫1x2+1dx+∫2x−1(x2+1)2dx
Now ∫1x2+1dx=tan−1x is a well known integral and for the other one
We can break ∫2x−1(x2+1)2dx=∫2x(x2+1)2dx−∫1(x2+1)2dx
For first term, let us assume u=x2+1, then du=2xdx= ∫2x(x2+1)2dx=∫duu2=−1u=−1x2+1
and for second term let us assume x=tanv then
∫1(x2+1)2dx=∫sec2vsec4vdv=∫cos2vdv
= 12∫(1+cos2v)dv=v2+14sin2v
= tan−1x2+12sinvcosv
= tan−1x2+12x√x2+1×1√x2+1
= tan−1x2+12xx2+1
Hence, ∫2x−1(x2+1)2dx
= −1x2+1+tan−1x2+12xx2+1+c
= 12(tan−1x)+x−2x2+1+c
