Question

If $f:R\to R$ is invertible function such that $\int f\left(x\right)dx=g\left(x\right)$,then $\int f^{-1}\left(x\right)dx$ is equal to (assuming c as arbitrary constant)

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

Answer 1

The correct option is B

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

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

Conceptual.