Question

Solve the following

$\int \: \frac{dx}{x \sqrt{ {x}^{15} - 1} }$
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Answer 1

$\frac{2}{15} tan^{ - 1} \sqrt{ {x}^{15} - 1} \: \: + c$

Step-by-step explanation:

$Given, \\ \int \frac{dx}{x \sqrt{ {x}^{15} - 1} } \\ = \int \frac{ {x}^{14}dx }{ {x}^{15} \sqrt{ {x}^{15} - 1} } \\ let \: \: \: y = \sqrt{ {x}^{15} - 1} \\ {y}^{2} = {x}^{15} - 1 \\ {y}^{2} + 1 = {x}^{15} \\ Differentiating \: \: with \: \: respect \: \: to \: \: x, \\ 2ydy = 15 {x}^{14} dx \\ {x}^{14} dx = \frac{2ydy}{15} \\ putting \: \: values\: \\ \int \frac{2ydy}{15( {y}^{2} + 1 )y} \\ = \frac{2}{15} \int \frac{dy}{ {y}^{2} + 1} \\ = \frac{2}{15} tan^{ - 1}y \: \: + c \\ = \frac{2}{15} tan^{ - 1} \sqrt{ {x}^{15} - 1} \: \: + c$