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Question Number 47743 by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

\int \frac{d x}{x^{n} - x}

t h e p r o b l e m s i h a v e p o s t e d a r e t r i c k y \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

\int \frac{d x}{x (x^{n - 1} - 1)}

\int \frac{x^{n - 2} d x}{x^{n - 1} (x^{n - 1} - 1)}

t = x^{n - 1} d t = (n - 1) x^{n - 2} d x

\int \frac{d t}{(n - 1) (t) (t - 1)}

\frac{1}{n - 1} \int \frac{t - (t - 1)}{t (t - 1)} d t

\frac{1}{n - 1} [\int \frac{d t}{t - 1} - \int \frac{d t}{t}]

\frac{1}{n - 1} [l n (\frac{t - 1}{t})] + c

\frac{1}{n - 1} [l n (\frac{x^{n - 1} - 1}{x^{n - 1}})] + c

Answered by ajfour last updated on 14/Nov/18

l e t t = \frac{1}{x} \Rightarrow d t = - \frac{d x}{x^{2}}

d x = - \frac{d t}{t^{2}}

I = \int \frac{(- \frac{d t}{t^{2}})}{\frac{1}{t^{n}} - \frac{1}{t}} = - \int \frac{t^{n - 2} d t}{1 - t^{n - 1}}

= \frac{1}{n - 1} \int \frac{- (n - 1) t^{n - 2} d t}{1 - t^{n - 1}}

I = \frac{1}{n - 1} \ln ∣ 1 - \frac{1}{t^{n - 1}} ∣ + c .

Commented by rahul 19 last updated on 14/Nov/18

Nice �� Is there any particular reason why u substitute t=1/x?

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

e x c e l l e n t \dots