$\int \frac{d x}{x (x^{5} + 1)}$ is equal to:

\frac{1}{5} l o g x^{5} (x^{5} + 1) + c

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\frac{1}{5} l o g [\frac{(x^{5} + 1)}{x^{5}}] + c

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\frac{1}{5} l o g [\frac{x^{5}}{(x^{5} + 1)}] + c

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None of these

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Solution

The correct option is B $\frac{1}{5} l o g [\frac{x^{5}}{(x^{5} + 1)}] + c$
$I = \int \frac{d x}{x (x^{5} + 1)} = \int \frac{d x}{x^{6} (1 + x^{- 5})}$ put $(1 + x^{- 5}) = t \Rightarrow - 5 x^{- 6} d x = d t$
$\Rightarrow I = \int \frac{x^{- 6} d x}{(1 + x^{- 5})} = - \frac{1}{5} \int \frac{d t}{t} = - \frac{1}{5} l o g t + c$
$= - \frac{1}{5} l o g (1 + x^{- 5}) + c = - \frac{1}{5} l o g [\frac{x^{5} + 1}{x^{5}}] + c$
$= \frac{1}{5} l o g {[\frac{x^{5} + 1}{x^{5}}]}^{- 1} + c = \frac{1}{5} l o g [\frac{x^{5}}{x^{5} + 1}] + c$

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