What is the integral of 1-x^{-0.52} ?

The integral of 1-x^{-0.52} is x-2.08333…x^{0.48}+C

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integral of 1-x^{-0.52}

\int 1 - x^{- 0.52} d x

x - 2.08333 \dots x^{0.48} + C

Solution steps

\int 1 - x^{- 0.52} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x - \int x^{- 0.52} d x

\int 1 d x = x

\int x^{- 0.52} d x = 2.08333 \dots x^{0.48}

= x - 2.08333 \dots x^{0.48}

Add a constant to the solution

= x - 2.08333 \dots x^{0.48} + C

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What is the integral of 1-x^{-0.52} ?
The integral of 1-x^{-0.52} is x-2.08333…x^{0.48}+C