_ ^ x^5 ;dx (1 + x^3 ) = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^5}\;dx}}{{\sqrt {(1 + {x^3})} }} = } $

A
$\frac{2}{3}\sqrt {(1 + {x^3})} ({x^3} + 2)$
B
$\frac{2}{9}\sqrt {(1 + {x^3})} ({x^3} - 4)$
C
$\frac{2}{9}\sqrt {(1 + {x^3})} ({x^3} + 4)$
✓
$\frac{2}{9}\sqrt {(1 + {x^3})} ({x^3} - 2)$

✓

Answer

Correct option: D.

$\frac{2}{9}\sqrt {(1 + {x^3})} ({x^3} - 2)$

d
(d) Here ${x^5} = {x^3}{x^2}$ and differential coefficient of ${x^3}$ is $3{x^{2}}$ In order to remove fractional powers, we put
$1 + {x^3} = {t^2} \Rightarrow 3{x^2}dx = 2t.\,dt\,;$ Also ${x^3} = {t^2} - 1$
$I = \int_{}^{} {\frac{{({t^2} - 1)}}{t}\left( {\frac{2}{3}t\,dt} \right) = \frac{2}{3}\int_{}^{} {({t^2} - 1)\,dt} } $
$ = \frac{2}{3}\left( {\frac{{{t^3}}}{3} - t} \right) = \frac{2}{9}t\,({t^2} - 3)$= $\frac{2}{9}\sqrt {(1 + {x^3})} \,({x^3} - 2)$ $(\because \,{t^2} = 1 + {x^3})$

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