Question

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

Answer 1

\[\text{ Let  I } = \int\frac{dx}{\sqrt{x^2 - a^2}}\]

\[\text{ Putting  x} = a \tan \theta\]

\[ \Rightarrow dx = a \sec^2  \text{ θ   dθ }\]

\[ \therefore I = \int\frac{a \cdot se c^2\text{ θ   dθ }}{\sqrt{a^2 \tan^2 \theta + a^2}}\]

\[ = \int\frac{a \sec^2 \theta \cdot d\theta}{a\sqrt{1 + \tan^2 \theta}}\]

\[ = \int\frac{\sec^2 \theta \cdot \text{    dθ }}{\sec\theta}\]

\[ = \int\sec\theta \cdot d\theta\]

\[ = \int\sec\theta \cdot d\theta\]

\[ = \text{ ln } \left| \sec\theta + \tan\theta \right| + C\]

\[ = \text{ ln }\left| \sec\theta + \sqrt{\sec^2 \theta - 1} \right| + C\]

\[ = \text{ ln }\left| \frac{x}{a} + \sqrt{\frac{x^2}{a^2} - 1} \right| + C\]

\[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| - \ln a + C\]

\[ = \text{ ln }\left| x + \sqrt{x^2 - a^2} \right| + C'\]

\[\text{ where C' = C -  ln  a }\]