_ ^ dx x( x^n + 1) = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{x({x^n} + 1)}} = } $

A
$n\log \frac{{{x^n}}}{{{x^n} + 1}} + c$
B
$n\log \frac{{{x^n} + 1}}{{{x^n}}} + c$
✓
$\frac{1}{n}\log \frac{{{x^n}}}{{{x^n} + 1}} + c$
D
$\frac{1}{n}\log \frac{{{x^n} + 1}}{{{x^n}}} + c$

✓

Answer

Correct option: C.

$\frac{1}{n}\log \frac{{{x^n}}}{{{x^n} + 1}} + c$

c
(c) Put ${x^n} = t \Rightarrow n{x^{n - 1}}dx = dt$
$ \Rightarrow \frac{{n{x^n}}}{x}\,dx = dt \Rightarrow \frac{1}{x}\,dx = \frac{{dt}}{{nt}},$ then it reduces to
$\int_{}^{} {\frac{{dt}}{{nt(t + 1)}}} = \frac{1}{n}\left[ {\int_{}^{} {\frac{{dt}}{{t(t + 1)}}} } \right]$
$ = \frac{1}{n}\left[ {\int_{}^{} {\frac{1}{t}\,dt - \int_{}^{} {\frac{1}{{t + 1}}\;dt} } } \right] = \frac{1}{n}\log \frac{{{x^n}}}{{{x^n} + 1}} + c$.

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