just found from nishant sir's bookmark..
http://www.targetiit.com/iit-jee-forum/posts/e-x-2-1849.html
Let t=f(z) Then ∫xyf−1(t)dt=∫f−1(x)f−1(y)zf,(z)dz =(zf(z))∣f−1(x)f−1(y)−∫f−1(x)f−1(y)f(z)dz (integrating by parts)
So, we get ∫xyf−1(t)dt=yf−1(y)−xf−1(x)−∫f−1(x)f−1(y)f(t)dt
Substituting x=f(a),y=f(b) we get
just found from nishant sir's bookmark..
http://www.targetiit.com/iit-jee-forum/posts/e-x-2-1849.html