Question

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

• A. $x\cdot f\left(x\right)-g\left(f^{-1}\right(x\left)\right)$
• B. $x{f}^{-1}\left(x\right)-g\left(f^{-1}\right(x\left)\right)$
• C. $x{f}^{-1}\left(x\right)-g\left(f\right(x\left)\right)$
• D. $x\cdot f^{-1}\left(x\right)$

Answer 1

The correct option is A $x\cdot f\left(x\right)-g\left(f^{-1}\right(x\left)\right)$

Now lets use integration by parts:

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

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

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

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

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