intx/(1+x)(1+x^2)dx=intx/(1+x)dx+intx^3/(1+x)dx
let
I_1=intx/(1+x)dx=int(1+x-1)/(1+x)dx=int(1-1/(1+x))dx
=int1dx-int1/(1+x)dx=x-ln(1+x)
I_2=intx^3/(1+x)dx=int(x^3+1-1)/(1+x)dx
=int(x^3+1)/(x+1)dx-int1/(1+x)dx
=int(x^2-x+1)dx-int1/(1+x)dx
=x^3/3-x^2/2+x-ln(x+1)
Now,
intx/(1+x)(1+x^2)dx=I_1+I_2
I_1=x-ln(x+1)
I_2=x^3/3-x^2/2+x-ln(x+1)
Thus,
intx/(1+x)(1+x^2)dx=x-ln(x+1)+x^3/3-x^2/2+x-ln(x+1)
Rearranging
intx/(1+x)(1+x^2)dx=x^3/3-x^2/2+2x-2ln(x+1)+C, C in RR