Let x2=t⇒2xdx=dt
∴∫2x(x2+1)(x2+3)dx=∫dt(t+1)(t+3) .........(1)
Let 1(t+1)(t+3)=A(t+1)+B(t+3)
⇒1=A(t+3)+b(t+1) ......... (1)
Substituting t=−3 and t=−1 in equation (1) we obtain
A=12 and B=−12
∴1(t+1)(t+3)=12(t+1)−12(t+3)
⇒∫2x(x2+1)(x2+3)dx=∫{12(t+1)−12(t+3)}dt
=12log|(t+1)|−12log|t+3|+C
=12log∣∣∣t+1t+3∣∣∣+C
=12log∣∣∣x2+1x2+3∣∣∣+C