_ ^ x^3 - 1 x^3 + x dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^3} - 1}}{{{x^3} + x}}dx = } $

A
$x - \log x + \frac{1}{2}\log ({x^2} + 1) + {\tan ^{ - 1}}x + c$
✓
$x - \log x + \log \sqrt {{x^2} + 1} - {\tan ^{ - 1}}x + c$
C
$x + \log x + \log \sqrt {{x^2} + 1} + {\tan ^{ - 1}}x + c$
D
None of these

✓

Answer

Correct option: B.

$x - \log x + \log \sqrt {{x^2} + 1} - {\tan ^{ - 1}}x + c$

b
(b) $\int_{}^{} {\frac{{{x^3} - 1}}{{{x^3} + x}}\,dx} = \int_{}^{} {\frac{{{x^3}}}{{x({x^2} + 1)}}\,dx - \int_{}^{} {\frac{1}{{x({x^2} + 1)\,}}} } \,dx$
$ = \int_{}^{} {\frac{{{x^2}}}{{{x^2} + 1}}\,dx} - \int_{}^{} {\left( {\frac{1}{x} - \frac{x}{{{x^2} + 1}}} \right)\,dx} $
$ = \int_{}^{} {\left( {1 - \frac{1}{{{x^2} + 1}}} \right)\,dx} - \int_{}^{} {\frac{1}{x}\,dx} + \int_{}^{} {\frac{x}{{{x^2} + 1}}\,dx} $
$ = x - {\tan ^{ - 1}}x - \log x + \log \sqrt {{x^2} + 1} + c.$

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