If $\displaystyle\int f(x)dx=g(x),$ then $\displaystyle\int f^{-1}(x)dx=$ _____________$+$c.

Now lets use integration by parts:

$\int { f^{ -1 }\left( x \right) dx } =xf^{ -1 }\left( x \right) -\int { x } \frac { d }{ dx } \left[ f^{ -1 }\left( x \right) \right] dx$

Now in the second integral, lets use substitution $x=f(u)$ ;

$\\ \int { f^{ -1 }\left( x \right) dx } =xf^{ -1 }\left( x \right) -\int { f\left( u \right) } d\left[ f^{ -1 }\left( f\left( u \right) \right) \right] \\ \int { f^{ -1 }\left( x \right) dx } =xf^{ -1 }\left( x \right) -g\left( u \right) +C$

But $u=f^{ -1 }\left( x \right)$,

$\therefore \int { f^{ -1 }\left( x \right) dx } =xf^{ -1 }\left( x \right) -g\left( f^{ -1 }\left( x \right) \right) +C$