Question

\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}} \text{ dx}\]

Answer 1

\[\text{ We  have,} \]
\[I = \int\frac{x dx}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}}\]
\[\text{ Putting  x}^2 = t\]
\[ \Rightarrow 2x \text{ dx } = dt\]
\[ \Rightarrow x \text{ dx} = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\frac{dt}{\left( t + 4 \right) \sqrt{t + 9}}\]
\[\text{ Again Putting  t} + 9 = u^2 \]
\[ \Rightarrow dt = 2u\text{  du }\]
\[ \therefore I = \frac{1}{2}\int\frac{2u \text{ du}}{\left( u^2 - 9 + 4 \right) u}\]
\[ = \int\frac{du}{u^2 - 5}\]
\[ = \int\frac{du}{u^2 - \left( \sqrt{5} \right)^2}\]
\[ = \frac{1}{2\sqrt{5}} \text{ log } \left| \frac{u - \sqrt{5}}{u + \sqrt{5}} \right| + C\]
\[ = \frac{1}{2\sqrt{5}} \text{ log } \left| \frac{\sqrt{t + 9} - \sqrt{5}}{\sqrt{t + 9} + \sqrt{5}} \right| + C\]
\[ = \frac{1}{2\sqrt{5}} \text{ log} \left| \frac{\sqrt{x^2 + 9} - \sqrt{5}}{\sqrt{x^2 + 9} + \sqrt{5}} \right| + C\]