Question

$\int \frac{1}{\left(2x^2+3\right)\sqrt{x^2-4}}dx$

Answer 1

​$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{\left(2x^2+3\right)\sqrt{x^2-4}} \\ \mathrm{Putting}x=\frac{1}{t} \\ \Rightarrow dx=-\frac{1}{t^2}dt \\ \therefore I=\int \frac{-{\displaystyle \frac{1}{t^2}}dt}{\left({\displaystyle \frac{2}{t^2}+3}\right)\sqrt{{\displaystyle \frac{1}{t^2}-4}}} \\ =\int \frac{-{\displaystyle \frac{1}{t^2}}dt}{{\displaystyle \frac{\left(2+3t^2\right)}{t^2}\times \frac{\sqrt{1-4t^2}}{t}}} \\ =-\int \frac{tdt}{\left(2+3t^2\right)\sqrt{1-4t^2}} \\ \mathrm{Again}\mathrm{Putting}1-4t^2=u^2 \\ \Rightarrow -8tdt=2udu \\ \Rightarrow tdt=-\frac{u}{4}du \\ \therefore I=\frac{1}{4}\int \frac{udu}{\left[2+3\left({\displaystyle \frac{1-u^2}{4}}\right)\right]u} \\ =\frac{1}{4}\int \frac{4du}{\left[8+3-3u^2\right]} \\ =\int \frac{du}{11-3u^2} \\ =\frac{1}{3}\int \frac{du}{{\displaystyle \frac{11}{3}-u^2}} \\ =\frac{1}{3}\int \frac{du}{{\left(\sqrt{{\displaystyle \frac{11}{3}}}\right)}^2-u^2} \\ =\frac{1}{3}\times \frac{1}{2\times {\displaystyle \frac{\sqrt{11}}{\sqrt{3}}}}\log \left|\frac{{\displaystyle \frac{\sqrt{11}}{\sqrt{3}}+u}}{{\displaystyle \frac{\sqrt{11}}{\sqrt{3}}-u}}\right|+C \\ =\frac{1}{2\sqrt{33}}\log \left|\frac{\sqrt{11}+\sqrt{3}u}{\sqrt{11}-\sqrt{3}u}\right|+C \\ =\frac{1}{2\sqrt{33}}\log \left|\frac{\sqrt{11}+\sqrt{3}\sqrt{1-4t^2}}{\sqrt{11}-\sqrt{3}\sqrt{1-4t^2}}\right|+C \\ =\frac{1}{2\sqrt{33}}\log \left|\frac{\sqrt{11}+\sqrt{3-12t^2}}{\sqrt{11}-\sqrt{3-12t^2}}\right|+C \\ =\frac{1}{2\sqrt{33}}\log \left|\frac{\sqrt{11}+\sqrt{3-{\displaystyle \frac{12}{x^2}}}}{\sqrt{11}-\sqrt{3-{\displaystyle \frac{12}{x^2}}}}\right|+C \\ =\frac{1}{2\sqrt{33}}\log \left|\frac{\sqrt{11}x+\sqrt{3x^2-12}}{\sqrt{11}x-\sqrt{3x^2-12}}\right|+C \end{gathered}$$