Mathematics

Integrate:
$$\displaystyle\int{x\sqrt{{{x}^{2}}+2}}dx$$


SOLUTION

Consider the given integral.

$$I=\int{x\sqrt{{{x}^{2}}+2}}dx$$

 Let $$t={{x}^{2}}+2$$

$$ \dfrac{dt}{dx}=2x+0 $$

$$ \dfrac{dt}{2}=xdx $$

 Therefore,

$$ I=\dfrac{1}{2}\int{\sqrt{t}}dt $$

$$ I=\dfrac{1}{2}\left( \dfrac{{{t}^{3/2}}}{\dfrac{3}{2}} \right)+C $$

$$ I=\dfrac{{{t}^{3/2}}}{3}+C $$

 On putting the value of $$t$$, we get

$$I=\dfrac{{{\left( {{x}^{2}}+2 \right)}^{3/2}}}{3}+C$$

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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