_ ^ x^4 (x - 1)( x^2 + 1) dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^4}}}{{(x - 1)({x^2} + 1)}}dx = } $

✓
$\frac{{x(x + 2)}}{2} + \frac{{\log (x - 1)}}{2} - \frac{{\log ({x^2} + 1)}}{4} - \frac{{{{\tan }^{ - 1}}x}}{2} + c$
B
$\frac{{x(x + 2)}}{2} + \frac{{\log (x - 1)}}{2} + \frac{{\log ({x^2} + 1)}}{4} - \frac{{{{\tan }^{ - 1}}x}}{2} + c$
C
$\frac{{x(x + 2)}}{2} + \frac{{\log (x - 1)}}{2} + \frac{{\log ({x^2} + 1)}}{4} + \frac{{{{\tan }^{ - 1}}x}}{2} + c$
D
None of these

✓

Answer

Correct option: A.

$\frac{{x(x + 2)}}{2} + \frac{{\log (x - 1)}}{2} - \frac{{\log ({x^2} + 1)}}{4} - \frac{{{{\tan }^{ - 1}}x}}{2} + c$

a
(a) $\int_{}^{} {\frac{{{x^4}}}{{(x - 1)({x^2} + 1)}}dx} = \int_{}^{} {\frac{{{x^4} - 1}}{{(x - 1)({x^2} + 1)}} + \int_{}^{} {\frac{1}{{(x - 1)({x^2} + 1)}}dx} } $
$ = \int_{}^{} {\frac{{(x + 1)(x - 1)({x^2} + 1)}}{{(x - 1)({x^2} + 1)}}} \,dx + \int_{}^{} {\frac{{dx}}{{(x - 1)({x^2} + 1)}}} $
$ = \int_{}^{} {(x + 1)\,dx} + \int_{}^{} {\frac{{dx}}{{(x - 1)({x^2} + 1)}}} $
$ = \frac{{{x^2}}}{2} + x + \left[ {\frac{1}{2}\log (x - 1) - \frac{1}{4}\log ({x^2} + 1) - \frac{1}{2}{{\tan }^{ - 1}}x} \right] + c$.

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