Question

\[\int\log \left( x + \sqrt{x^2 + a^2} \right) \text{ dx}\]

Answer 1

\[Let I = \int {1_{II} \cdot \log}_{} \left( x + \sqrt{x^2_I + a^2} \right)\text{ dx}\]
\[ = \text{ log} \left( x + \sqrt{x^2 + a^2} \right)\int1 \text{ dx} - \int\left[ \frac{d}{dx}\left\{ \text{ log }\left( x + \sqrt{x^2 + a^2} \right) \right\}\int1\text{ dx} \right]\]
\[ = \text{ log} \left( x + \sqrt{x^2 + a^2} \right) \cdot x - \int\left( \frac{1}{x + \sqrt{x^2 + a^2}} \right) \times \left( 1 + \frac{1 \times 2x}{2\sqrt{x^2 + a^2}} \right) \cdot x \cdot dx\]
\[ = \text{ log }\left( x + \sqrt{x^2 + a^2} \right) \cdot x - \int\frac{x}{\sqrt{x^2 + a^2}}dx\]
\[\text{Putting   x}^2 + a^2 = t\ \text{in the second integra}l\]
\[ \Rightarrow\text{  2x  dx = dt}\]
\[ \Rightarrow x \text{ dx }= \frac{dt}{2}\]
\[ \therefore I = x \cdot \text{ log } \left( x + \sqrt{x^2 + a^2} \right) - \frac{1}{2}\int\frac{1}{\sqrt{t}}dt\]
\[ = x \cdot \text{ log} \left( x + \sqrt{x^2 + a^2} \right) - \frac{1}{2}\int t^{- \frac{1}{2}} dt\]
\[ = x \cdot \text{ log } \left( x + \sqrt{x^2 + a^2} \right) - \frac{1}{2} \left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]
\[ = x \cdot \text{ log }\left( x + \sqrt{x^2 + a^2} \right) - \sqrt{t} + C\]
\[ = x \cdot \text{ log }\left( x + \sqrt{x^2 + a^2} \right) - \sqrt{x^2 - a^2} + C.............. \left[ \because t = x^2 + a^2 \right]\]