The correct option is A I=x4a2(x2+a2)2+34a2{x2a2(x2+a2)+12a3tan−1(xa)}+C
Let I=∫dx(x2+a2)3 ....(i)
and I1=∫1(x2+a2)2dx ......(ii)
I1=∫1(x2+a2)2.1dx
=1(x2+a2)2.x−∫−2(2x)(x2+a2)3xdx
=x(x2+a2)2+4∫x2+a2−a2(x2+a2)3dx
=x(x2+a2)2+4∫1(x2+a2)2dx−4a2∫dx(x2+a2)3
⟹I1=x(x2+a2)2+4I1−4a2.I [using equations (i) and (ii)]
⟹4a2.I=x(x2+a2)2+3I1
⟹I=x4a2(x2+a2)2+34a2I1 (iii)
⟹I=x4a2(x2+a2)2+34a2{x2a2(x2+a2)+12a3tan−1(xa)}+C