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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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