Mathematics

# Integrate:$\displaystyle\int{\left( x+1 \right)\sqrt{3-2x-{{x}^{2}}}}dx =$

##### SOLUTION

Consider the given integral.

$I=\int{\left( x+1 \right)\sqrt{3-2x-{{x}^{2}}}}dx$

Let $t=3-2x-{{x}^{2}}$

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

$-\dfrac{dt}{2}=\left( x+1 \right)dx$

Therefore,

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

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

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

On putting the value of $t$, we have

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

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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