Mathematics

# $\displaystyle\int { \dfrac { { x }^{ 5 } }{ \sqrt { 1+{ x }^{ 3 } } } dx }$ is equal to

$\dfrac { 2 }{ 9 } \sqrt { 1+{ x }^{ 2 } } \left( { x }^{ 3 }-9 \right) +C$

##### SOLUTION
Let $I=\displaystyle\int { \dfrac { { x }^{ 5 } }{ \sqrt { 1+{ x }^{ 3 } } } dx } =\displaystyle\int { \dfrac { { x }^{ 3 }\cdot { x }^{ 2 } }{ \sqrt { 1+{ x }^{ 3 } } } dx }$
Put $1+{ x }^{ 3 }={ t }^{ 2 }$
$\Rightarrow 3{ x }^{ 2 }dx=2tdt$
$\Rightarrow { x }^{ 2 }dx=\dfrac { 2 }{ 3 } tdt$
Also, ${ x }^{ 3 }={ t }^{ 2 }-1$
$\therefore I=\dfrac { 2 }{ 3 } \displaystyle\int { \dfrac { \left( { t }^{ 2 }-1 \right) }{ t } tdt }$
$=\dfrac { 2 }{ 3 } \displaystyle\int { \left( { t }^{ 2 }-1 \right) dt }$
$=\dfrac { 2 }{ 3 } \left[ \dfrac { { t }^{ 3 } }{ 3 } -t \right] +C$
$=\dfrac { 2 }{ 3 } t\left[ \dfrac { { t }^{ 2 } }{ 3 } -1 \right] +C$
$\Rightarrow I=\dfrac { 2 }{ 9 } t\left( { t }^{ 2 }-3 \right) +C$
Put $t=\sqrt { 1+{ x }^{ 3 } }$
$\Rightarrow I=\dfrac { 2 }{ 9 } \sqrt { 1+{ x }^{ 3 } } \left( { x }^{ 3 }-2 \right) +C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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