Question

If $\int f\left(x\right)dx=g\left(x\right)$, then $\int f^{-1}\left(x\right)dx$ is
[Where $C$ is constant of integration]

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

Answer 1

Answer: (C)

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

$Q\int f\left(x\right)dx=g\left(x\right)then\int f^{-1}\left(x\right)dx=?$

Let $f^{-1}\left(x\right)=t$

$x=f\left(t\right)$

$dx=f^1\left(t\right)dt$

$\Rightarrow \int f^{-1}\left(x\right)dx=\int t.f^1\left(t\right)dt$

$=t\int f^1\left(t\right)dt-\int \{\frac{d}{dt}t\int f^1(t\left)dt\right\}dt$

$=tf\left(t\right)-\int f\left(t\right)dt$

$=tf\left(t\right)-g\left(t\right)$

$=f^{-1}\left(x\right).f\left(f^{-1}\right(x\left)\right)-g\left(f^{-1}\right(x\left)\right)+c$

$=f^{-1}\left(x\right).x-g\left(f^{-1}\right(x\left)\right)+c$

$=x{f}^{-1}\left(x\right)-g\left(f^{-1}\right(x\left)\right)+c$