_ ^ ( x^4 - x) ^ 1/4 x^5 ;dx is equal to - Vidyadip

MCQ

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

✓
$\frac{4}{{15}}{\left( {1 - \frac{1}{{{x^3}}}} \right)^{5/4}} + c$
B
$\frac{4}{5}{\left( {1 - \frac{1}{{{x^3}}}} \right)^{5/4}} + c$
C
$\frac{4}{{15}}{\left( {1 + \frac{1}{{{x^3}}}} \right)^{5/4}} + c$
D
None of these

✓

Answer

Correct option: A.

$\frac{4}{{15}}{\left( {1 - \frac{1}{{{x^3}}}} \right)^{5/4}} + c$

a
(a)$\int_{}^{} {\frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}}\,dx} = \int_{}^{} {\frac{1}{{{x^4}}}{{\left( {1 - \frac{1}{{{x^3}}}} \right)}^{1/4}}dx} $
$ = \frac{1}{3}\int_{}^{} {{t^{1/4}}dt} = \frac{4}{{15}}{t^{5/4}} + c = \frac{4}{{15}}{\left( {1 - \frac{1}{{{x^3}}}} \right)^{5/4}} + c$
$\left\{ {{\rm{Putting}}\,1 - \frac{1}{{{x^3}}} = t\,{\rm{and}}\,\frac{1}{{{x^4}}}\,dx = \frac{1}{3}\,dt} \right\}$

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