_ ^ x^5 1 + x^3 dx = - Vidyadip

MCQ

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

A
$\frac{2}{9}{(1 + {x^3})^{3/2}} + c$
B
$\frac{2}{9}{(1 + {x^3})^{3/2}} + \frac{2}{3}{(1 + {x^3})^{1/2}} + c$
✓
$\frac{2}{9}{(1 + {x^3})^{3/2}} - \frac{2}{3}{(1 + {x^3})^{1/2}} + c$
D
None of these

✓

Answer

Correct option: C.

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

c
(c) Put $1 + {x^3} = {t^2} \Rightarrow 3{x^2}dx = 2t\,dt$ and ${x^3} = {t^2} - 1$
So, $\int_{}^{} {\frac{{{x^5}}}{{\sqrt {1 + {x^3}} }}\,dx = \int_{}^{} {\frac{{{x^2}.{x^3}}}{{\sqrt {1 + {x^3}} }}\,dx} } $
$ = \frac{2}{3}\int_{}^{} {\frac{{({t^2} - 1)\,.\,t\,dt}}{t} = \frac{2}{3}\int_{}^{} {({t^2} - 1)\,dt = \frac{2}{3}\left[ {\frac{{{t^3}}}{3} - t} \right]} } + c$
$ = \frac{2}{3}\left[ {\frac{{{{(1 + {x^3})}^{32}}}}{3} - {{(1 + {x^3})}^{12}}} \right] + c$.

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