Question

If $f$ is an invertible function such that $\int f\left(x\right)dx=g\left(x\right)$ for some function $g$, then $\int f^{-1}\left(x\right)dx$ is equal to

• A. $x{f}^{-1}\left(x\right)-g\left(f^{-1}\right(x\left)\right)+C$, where $C$ is a constant.
  
• B. $f^{-1}\left(x\right)-xg\left(f^{-1}\right(x\left)\right)+C$, where $C$ is a constant.
• C. $(x-1)f^{-1}\left(x\right)+xg\left(f^{-1}\right(x\left)\right)+C$, where $C$ is a constant.
• D. $\frac{x^2}{2}{f}^{-1}\left(x\right)+g\left(f^{-1}\right(x\left)\right)+C$, where $C$ is a constant.

Answer 1

The correct option is A $x{f}^{-1}\left(x\right)-g\left(f^{-1}\right(x\left)\right)+C$, where $C$ is a constant.


We know that,
$f\left(x\right)=g^{\prime }\left(x\right)$
$I=\int f^{-1}\left(x\right)dx$
Assuming $f^{-1}\left(x\right)=t$
$f\left(t\right)=x\Rightarrow f^{\prime }\left(t\right)dt=dx\Rightarrow g^{\prime \prime }\left(t\right)dt=dx$

$I=\int f^{-1}\left(x\right)dx\Rightarrow I=\int t{g}^{\prime \prime }\left(t\right)dt\Rightarrow I=t{g}^{\prime }\left(t\right)-\int g^{\prime }\left(t\right)dt\Rightarrow I=t{g}^{\prime }\left(t\right)-g\left(t\right)+C\Rightarrow I=f^{-1}\left(x\right)f\left(t\right)-g\left(f^{-1}\right(x\left)\right)+C\Rightarrow I=x{f}^{-1}\left(x\right)-g\left(f^{-1}\right(x\left)\right)+C$