Question

$\int \frac{1}{\sqrt{a^2+b^2{x}^2}}dx$

Answer 1

$$\begin{gathered} \int \frac{dx}{\sqrt{a^2+b^2{x}^2}} \\ =\int \frac{dx}{\sqrt{b^2\left({\displaystyle \frac{a^2}{b^2}}+x^2\right)}} \\ =\frac{1}{b}\int \frac{dx}{\sqrt{x^2+{\left({\displaystyle \frac{a}{b}}\right)}^2}} \\ =\frac{1}{b}\log \left|x+\sqrt{x^2+\frac{a^2}{b^2}}\right|+C \\ =\frac{1}{b}\left[\log \left|x+\frac{\sqrt{b^2{x}^2+a^2}}{b}\right|\right]+C \\ =\frac{1}{b}\left[\log \left|\frac{bx+\sqrt{b^2{x}^2+a^2}}{b}\right|\right]+C \\ =\frac{1}{b}\left[\log \left|bx+\sqrt{b^2{x}^2+a^2}\right|-\log b\right]+C \\ =\frac{1}{b}\log \left|bx+\sqrt{b^2{x}^2+a^2}\right|-\frac{\log b}{b}+C \\ \mathrm{let}C-\frac{\log b}{b}=C\text{'} \\ =\frac{1}{b}\log \left|bx+\sqrt{b^2{x}^2+a^2}\right|+C\text{'} \end{gathered}$$