Mathematics

Integral of $$\displaystyle f\left ( x \right )=\sqrt{1+x^{2}}$$ with respect to $$\displaystyle x^{2}$$ is


ANSWER

$$\displaystyle \frac{2}{3}\left ( 1+x^{2} \right )^{\tfrac 32}+k$$


SOLUTION
Let $$\displaystyle I=\int { f\left( x \right) d\left( { x }^{ 2 } \right)  } =\int { 2x\sqrt { 1+{ x }^{ 2 } } dx } $$

Substitute $$t=\sqrt { 1+{ x }^{ 2 } } \Rightarrow dt=2xdx$$

$$\displaystyle \therefore I=\int { \sqrt { t } dt } =\dfrac { 2 }{ 3 } { t }^{ \dfrac { 3 }{ 2 }  }+c$$

$$\displaystyle =\dfrac { 2 }{ 3 } { \left( 1+{ x }^{ 2 } \right)  }^{ \dfrac { 3 }{ 2 }  }+k$$
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Single Correct Medium Published on 17th 09, 2020
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