Question

$\int \frac{1}{\sqrt{7-3x-2x^2}}dx$

Answer 1

$$\begin{gathered} \int \frac{dx}{\sqrt{7-3x-2x^2}} \\ =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{{\displaystyle \frac{7}{2}}-{\displaystyle \frac{3}{2}}x-x^2}} \\ =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{{\displaystyle \frac{7}{2}}-\left(x^2-{\displaystyle \frac{3}{2}}x\right)}} \\ =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{{\left({\displaystyle \frac{\sqrt{7}}{\sqrt{2}}}\right)}^2-\left(x^2+{\displaystyle \frac{3}{2}}x+{\left({\displaystyle \frac{3}{4}}\right)}^2-{\left({\displaystyle \frac{3}{4}}\right)}^2\right)}} \\ =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{{\left({\displaystyle \frac{\sqrt{7}}{\sqrt{2}}}\right)}^2-{\left(x+{\displaystyle \frac{3}{4}}\right)}^2+{\displaystyle \frac{9}{16}}}} \\ =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{{\displaystyle \frac{7}{2}}+{\displaystyle \frac{9}{16}}-{\left(x+{\displaystyle \frac{3}{4}}\right)}^2}} \\ =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{{\displaystyle \frac{56+9}{16}}-{\left(x+{\displaystyle \frac{3}{4}}\right)}^2}} \\ =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{{\left({\displaystyle \frac{\sqrt{65}}{4}}\right)}^2-{\left(x+{\displaystyle \frac{3}{4}}\right)}^2}} \\ =\frac{1}{\sqrt{2}}{\sin }^{-1}\left[\frac{x+{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{\sqrt{65}}{4}}}\right]+C \\ =\frac{1}{\sqrt{2}}{\sin }^{-1}\left[\frac{4x+3}{\sqrt{65}}\right]+C \end{gathered}$$