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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