Question

Write a value of $\int \sqrt{x^2-9}dx$.

Answer 1

$$\begin{gathered} \int \sqrt{x^2-9}dx \\ =\int \sqrt{x^2-3^2}dx \\ =\frac{x}{2}\sqrt{x^2-3^2}-\frac{3^2}{2}\ln \left|x+\sqrt{x^2-3}\right|+C\left(\because \sqrt{x^2-a^2}=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln \left|x+\sqrt{x^2+a^2}\right|+C\right) \\ =\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}\ln \left|x+\sqrt{x^2-9}\right|+C \end{gathered}$$