Question

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

Answer 1

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