What is the integral of f^{-1}(x) ?

The integral of f^{-1}(x) is (x^2)/(2f)+C

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integral of f^{-1}(x)

\int f^{- 1} (x) d x

\frac{x ^{2}}{2 f} + C

Solution steps

\int f^{- 1} x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= f^{- 1} \cdot \int x d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{f} \cdot \int x d x

Apply the Power Rule

= \frac{1}{f} \cdot \frac{x ^{2}}{2}

Simplify

\frac{1}{f} \cdot \frac{x ^{2}}{2} : \frac{x ^{2}}{2 f}

= \frac{x ^{2}}{2 f}

Add a constant to the solution

= \frac{x ^{2}}{2 f} + C