Question

$\int \sqrt{1+x-2x^2}dx$

Answer 1

$$\begin{gathered} \int \sqrt{1+x-2x^2}dx \\ =\int \sqrt{2\left(\frac{1}{2}+\frac{x}{2}-x^2\right)}dx \\ =\sqrt{2}\int \sqrt{\frac{1}{2}-\left(x^2-\frac{x}{2}\right)}dx \\ =\sqrt{2}\int \sqrt{\frac{1}{2}-\left(x^2-\frac{x}{2}+\frac{1}{4^2}-\frac{1}{4^2}\right)}dx \\ =\sqrt{2}\int \sqrt{\frac{1}{2}+\frac{1}{16}-{\left(x-\frac{1}{4}\right)}^2}dx \\ =\sqrt{2}\int \sqrt{{\left(\frac{3}{4}\right)}^2-{\left(x-\frac{1}{4}\right)}^2}dx \\ =\sqrt{2}\left[\left(\frac{x-{\displaystyle \frac{1}{4}}}{2}\right)\sqrt{{\left(\frac{3}{4}\right)}^2-{\left(x-\frac{1}{4}\right)}^2}+\frac{9}{32}{\sin }^{-1}\left(\frac{x-{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{3}{4}}}\right)\right]+C\left[\because \int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{1}{2}{a}^2{\sin }^{-1}\frac{x}{a}+C\right] \\ =\left(\frac{4x-1}{8}\right)\sqrt{1+x-2x^2}+\frac{9\sqrt{2}}{32}{\sin }^{-1}\left(\frac{4x-1}{3}\right)+C \end{gathered}$$