Question

Integrate:

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

Answer 1

Consider the given integral.

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

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

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

$-\frac{dt}{2}=(x+1)dx$

Therefore,

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

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

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

On putting the value of $t$, we have

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

Hence, this is the answer.