The correct option is D 1a2−b2
Writing the LHS in partial fractions
By substituting x2=z
z(z+a2)(z+b2)=Az+a2+Bz+b2
We get the values of A and B
By comparing co-efficients we get
A=a2a2−b2 and B=−b2a2−b2
x2(x2+a2)(x2+b2)=a2(a2−b2)(x2+a2)−b2(a2−b2)(x2+b2)
=1a2−b2[a2x2+a2−b2x2+b2]