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
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