Integrate inverse of $f(x)$?

Question

If $f(x)= x^3+x$, then $\displaystyle\int_1^2f(x)\,dx + 2\int_1^5f^{-1}(2x)\;dx $ is -

My Approach: let $f^{-1}(2x)=t$

$\implies$ $f(t)=2x$

$\implies$ $f'(t)dt=2dx$

$\displaystyle\int_1^2f(x)\,dx + \int_1^2t f'(t)dt$

Using integration by-parts in second integral

$\displaystyle\int_1^2f(x)\,dx+ (tf(t))^2_1 - \int_1^2 f(t)dt$

So final Answer will be $18$

Is my Approach correct?

Answer 1

There is a nice handy formula for integration of inverse functions given by;

$\displaystyle\int f^{-1}(y)\,dy = x\cdot f^{-1}(x) - F\circ f^{-1}(x)+C$

Where $F $ is an antiderivative of $f$

Lemma: If $f(x)$ is a continuous and increasing function and $a\lt b$ then :

$$\displaystyle\int_a^bf(x)\,dx +\int_{f(a)}^{f(b)}f^{-1}(x)\,dx = bf(b)-af(a)$$

You can look at this wiki page for the proof with a picture

Can you now solve your problem??

EDIT: here is link to a .pdf of some examples of such inverse functions