Mathematics

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


ANSWER

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


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
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