Question

Solve
$I=\int \frac{1}{\sqrt{21+4x-4x^2}}dx$

Answer 1

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$\begin{array}{c}\Rightarrow \int \frac{1}{2\sqrt{-\left(x^2-x-\frac{21}{4}\right)}}dx\\ \Rightarrow \frac{1}{2}\int \frac{1}{\sqrt{-\left(x^2-2.x\cdot \frac{1}{2}+\frac{1}{4}-\frac{22}{4}\right)}}dx\\ \Rightarrow \frac{1}{2}\int \frac{1}{\sqrt{{\left(\sqrt{\frac{11}{2}}\right)}^2-{\left(x-\frac{1}{2}\right)}^2}}dx\\ \Rightarrow \frac{1}{2}\left[\frac{1}{\frac{\sqrt{11}}{\sqrt{2}}}{\sin }^{-1}\left[\frac{x-\frac{1}{2}}{\frac{\sqrt{11}}{\sqrt{2}}}\right]\right]\\ \Rightarrow \frac{1}{\sqrt{22}}{\sin }^{-1}\left(\frac{2x-1}{\sqrt{22}}\right)+C\end{array}$