_ ^ x^3 x^2 + 2 dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^3}}}{{\sqrt {{x^2} + 2} }}dx = } $

A
$\frac{1}{3}{({x^2} + 2)^{3/2}} + 2{({x^2} + 2)^{1/2}} + c$
✓
$\frac{1}{3}{({x^2} + 2)^{3/2}} - 2{({x^2} + 2)^{1/2}} + c$
C
$\frac{1}{3}{({x^2} + 2)^{3/2}} + {({x^2} + 2)^{1/2}} + c$
D
$\frac{1}{3}{({x^2} + 2)^{3/2}} - {({x^2} + 2)^{1/2}} + c$

✓

Answer

Correct option: B.

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

b
(b)$\int_{}^{} {\frac{{{x^3}}}{{\sqrt {{x^2} + 2} }}\,dx} = \int_{}^{} {\frac{{{x^2}.\,x}}{{\sqrt {{x^2} + 2} }}\,dx} $
Put ${x^2} + 2 = {t^2} \Rightarrow x\,dx = t\,dt$ and ${x^2} = {t^2} - 2,$ then it reduces to $\int_{}^{} {\frac{{({t^2} - 2)}}{t}\,t\,dt} = \int_{}^{} {({t^2} - 2)dt} $
$ = \frac{{{t^3}}}{3} - 2t + c = \frac{{{{({x^2} + 2)}^{32}}}}{3} - 2{({x^2} + 2)^{1/2}} + c.$

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