Let 2(1−x)(1+x2)=A(1−x)+Bx+C(1+x2)
⇒2=A(1+x2)+(Bx+C)(1−x)
⇒2=A+Ax2+Bx−Bx2+C+Cx
Equating the coefficient of x2,x, and constant term, we obtain
A−B=0
B−C=0
A+C=2
On solving these questions, we obtain
A=1,B=1, and C=1
∴2(1−x)(1+x2)=11−x+x+11+x2
⇒2(1−x)(1+x2)dx=∫11−xdx+∫x1+x2dx+∫11+x2dx
=−∫1x−1dx+12∫2x1+x2dx+∫11+x2dx
=−log|x−1|+12log|1+x2|+tan−1x+C