Question

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

Answer 1

\[\text{ We  have, }\]
\[I = \int \frac{x dx}{\left( x^2 + 4 \right) \sqrt{x^2 + 1}}\]
\[\text{ Putting}\ x^2 = t\]
\[ \Rightarrow 2x \text{ dx }= dt\]
\[ \Rightarrow x \text{ dx } = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int \frac{dt}{\left( t + 4 \right) \sqrt{t + 1}}\]
\[\text{ Again  Putting } t + 1 = p^2 \]
\[ \Rightarrow t = p^2 - 1\]
\[ \Rightarrow dt = 2p \text{ dp }\]
\[I = \frac{1}{2}\int \frac{2p \text{ dp }}{\left( p^2 - 1 + 4 \right)p}\]
\[ = \int \frac{dp}{p^2 + 3}\]
\[ = \int\frac{dp}{p^2 + \left( \sqrt{3} \right)^2}\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{p}{\sqrt{3}} \right) + C\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{\sqrt{t + 1}}{\sqrt{3}} \right) + C\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \sqrt{\frac{x^2 + 1}{3}} \right) + C\]