Mathematics

$$\displaystyle\int { \dfrac { { x }^{ n-1 } }{ { x }^{ 2n }+{ a }^{ 2 } } dx } $$ is equal to


ANSWER

$$\dfrac { n }{ a } \tan ^{ -1 }{ \left( \dfrac { { x }^{ n } }{ a } \right) } +C$$


SOLUTION
Let $$I=\displaystyle\int { \dfrac { { x }^{ n-1 } }{ { x }^{ 2n }+{ a }^{ 2 } } dx } $$
Put $${ x }^{ n }=t$$
$$\Rightarrow n{ x }^{ n-1 }dx=dt$$
Therefore, $$ I=\dfrac { 1 }{ n } \displaystyle\int { \dfrac { dt }{ { t }^{ 2 }+{ a }^{ 2 } }  } $$
$$=\dfrac { 1 }{ n } \cdot \dfrac { 1 }{ a } \tan ^{ -1 }{ \dfrac { t }{ a }  } +C$$
Put $$t={ x }^{ n }$$
$$\Rightarrow I=\dfrac { 1 }{ na } \tan ^{ -1 }{ \left( \dfrac { { x }^{ n } }{ a }  \right)  } +C$$
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Single Correct Medium Published on 17th 09, 2020
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