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

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

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integral of 1-t^{-0.52}

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

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

Solution steps

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

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 t - \int t^{- 0.52} d t

\int 1 d t = t

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

= t - 2.08333 \dots t^{0.48}

Add a constant to the solution

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

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