_ ^ 1 [ (x - 1) ^3 (x + 2) ^5 ] ^ 1/4 ;dx is equal to - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{{{[{{(x - 1)}^3}{{(x + 2)}^5}]}^{1/4}}}}\;dx} $ is equal to

✓
$\frac{4}{3}{\left( {\frac{{x - 1}}{{x + 2}}} \right)^{1/4}} + c$
B
$\frac{4}{3}{\left( {\frac{{x + 2}}{{x - 1}}} \right)^{1/4}} + c$
C
$\frac{1}{3}{\left( {\frac{{x - 1}}{{x + 2}}} \right)^{1/4}} + c$
D
$\frac{1}{3}{\left( {\frac{{x + 2}}{{x - 1}}} \right)^{1/4}} + c$

✓

Answer

Correct option: A.

$\frac{4}{3}{\left( {\frac{{x - 1}}{{x + 2}}} \right)^{1/4}} + c$

a
(a)$\int_{}^{} {\frac{1}{{{{[{{(x - 1)}^3}{{(x + 2)}^5}]}^{1/4}}}}} \,dx = \int_{}^{} {\frac{1}{{{{\left( {\frac{{x - 1}}{{x + 2}}} \right)}^{3/4}}{{(x + 2)}^2}}}} \,dx$
$ = \frac{1}{3}\int_{}^{} {\frac{1}{{{t^{3/4}}}}\,dt} $, $\left\{ {\because \,\,\,\frac{{x - 1}}{{x + 2}} = t \Rightarrow \frac{3}{{{{(x + 2)}^2}}}\,dx = dt} \right\}$
$ = \frac{1}{3}\left( {\frac{{{t^{1/4}}}}{{1/4}}} \right) + c = \frac{4}{3}{t^{1/4}} + c = \frac{4}{3}{\left( {\frac{{x - 1}}{{x + 2}}} \right)^{1/4}} + c$.

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