Question

Enter 1 for True and 0 for False $\int (x+3)\sqrt{3-4x-x^2}dx=\frac{-1}{3}{(3-4x-x^2)}^{3/2}+\frac{(x+2)\sqrt{3-4x-x^2}}{2}+\frac{7}{2}{\sin }^{-1}\frac{(x+2)}{\sqrt{7}}+C$

Answer 1

$\int (x+3)\sqrt{3-4x-x^2}dx$
$=\frac{-1}{2}\int (-2x-6)\sqrt{3-4x-x^2}dx$
$=\frac{-1}{2}\int (-2x-4)\sqrt{3-4x-x^2}dx+\int \sqrt{3-4x-x^2}dx$
Put $t=3-4x-x^2$ in the first integral.
$\Rightarrow dt=-4-2x$
$=\frac{-1}{2}\int \sqrt{t}dt+\int \sqrt{7-(4x+x^2+4)}dx$
$=\frac{-1}{3}{t}^{3/2}+\int \sqrt{{\left(\sqrt{7}\right)}^2-(x+2{)}^2}dx$
$=\frac{-1}{3}{(3-4x-x^2)}^{3/2}+\frac{(x+2)\sqrt{3-4x-x^2}}{2}+\frac{7}{2}{\sin }^{-1}\frac{(x+2)}{\sqrt{7}}+C$