Mathematics

# Prove that $\displaystyle \int \dfrac {f'(x)}{\sqrt [n]{f(x)}} dx = \dfrac {[f(x)]^{1 - \tfrac {1}{n}}}{1 - \dfrac {1}{n}} + C, n\neq 1$

##### SOLUTION
$I = \displaystyle \int \dfrac {f'(x)dx}{\sqrt [n]{f(x)}}$

Let $f(x) = t; f'(x) dx = dt$

$= \displaystyle \int \dfrac {1}{\sqrt [n]{t}} dt$

$= \displaystyle \int t^{-\tfrac {1}{n}} dt$

$= \dfrac {t^{-\tfrac {1}{n} + 1}}{-\dfrac {1}{n} + 1} + c$

$= \dfrac {[f(x)]^{1 - \tfrac {1}{n}}}{1 - \dfrac {1}{n}} + c$.

$\therefore \displaystyle \int \dfrac {f'(x)}{\sqrt [n]{f(x)}} dx = \dfrac {[f(x)]^{1 - \tfrac {1}{n}}}{1 - \dfrac {1}{n}} + C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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