What is the integral of (\sqrt[4]{1-u^{-1}})/(u^2) ?

The integral of (\sqrt[4]{1-u^{-1}})/(u^2) is 4/5 ((u-1)/u)^{5/4}+C

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integral of (\sqrt[4]{1-u^{-1}})/(u^2)

\int \frac{4 1 - u ^{- 1}}{u ^{2}} d u

\frac{4}{5} (\frac{u - 1}{u})^{\frac{5}{4}} + C

Solution steps

\int \frac{4 1 - u ^{- 1}}{u ^{2}} d u

Apply u-substitution

: \int 4 v d v

= \int 4 v d v

Apply the Power Rule

: \frac{4}{5} v^{\frac{5}{4}}

= \frac{4}{5} v^{\frac{5}{4}}

Substitute back

v = 1 - u^{- 1}

= \frac{4}{5} (1 - u^{- 1})^{\frac{5}{4}}

(1 - u^{- 1})^{\frac{5}{4}} = (\frac{u - 1}{u})^{\frac{5}{4}}

= \frac{4}{5} (\frac{u - 1}{u})^{\frac{5}{4}}

Add a constant to the solution

= \frac{4}{5} (\frac{u - 1}{u})^{\frac{5}{4}} + C

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What is the integral of (\sqrt[4]{1-u^{-1}})/(u^2) ?
The integral of (\sqrt[4]{1-u^{-1}})/(u^2) is 4/5 ((u-1)/u)^{5/4}+C