Question

\[\int e^x \sqrt{e^{2x} + 1} \text{ dx}\]

Answer 1

\[\text{ Let I } = \int e^x \sqrt{e^{2x} + 1} \text{ dx}\]

\[\text{ Putting}\ e^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[ \therefore I = \int \sqrt{t^2 + 1}\text{ dt}\]
\[ = \frac{t}{2}\sqrt{t^2 + 1} + \frac{1^2}{2}\text{ ln } \left| t + \sqrt{t^2 + 1} \right| + C \left[ \because \int\sqrt{x^2 + a^2}\text{ dx } = \frac{1}{2}x\sqrt{x^2 + a^2} + \frac{1}{2}\text{ ln }\left| x + \sqrt{x^2 + a^2} \right| + C \right] \]
\[ = \frac{e^x}{2} \sqrt{e^{2x} + 1} + \frac{1}{2}\text{ ln }\left| e^x + \sqrt{e^{2x} + 1} \right| + C \left( \because t = e^x \right)\]