Question

Solve
$\int (x-1)\sqrt{1-x-x^2}dx$

Answer 1

$\begin{array}{c}\int \left(x-1\right)\sqrt{1-x-x^2}dx\\ =\int \left(-\frac{1}{2}\left(-2x-1\right)-\frac{3}{2}\right)\sqrt{1-x-x^2}dx\\ =-\frac{1}{2}\int \left(-1-2x\right)\sqrt{1-x-x^2}dx-\frac{3}{2}\int \sqrt{1-x-x^2}dx\\ =-\frac{1}{3}{\left(1-x-x^2\right)}^{\frac{3}{2}}-\frac{3}{2}\int \sqrt{{\left(\frac{\sqrt{5}}{2}\right)}^2-{\left(x+\frac{1}{2}\right)}^2}dx\\ =-\frac{1}{3}{\left(1-x-x^2\right)}^{\frac{3}{2}}-\frac{3}{2}\left[\frac{x+\frac{1}{2}}{2}\sqrt{1-x-x^2}+\frac{5}{8}{\sin }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{5}}{2}}\right)\right]+C\\ =-\frac{1}{3}{\left(1-x-x^2\right)}^{\frac{3}{2}}-\frac{3}{2}\left\{\frac{\left(2x+1\right)}{4}\sqrt{1-x-x^2}+\frac{5}{8}{\sin }^{-1}\left(\frac{2x+1}{\sqrt{5}}\right)\right\}+C\\ Hence,solved.\end{array}$