Evaluate the following integrals:

Let $\mathrm{bx}=\mathrm{t}$ then $\mathrm{dt}=\mathrm{bdx}$ or $\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{b}}$

Hence, $\int \frac{1}{\sqrt{a^{2}+b^{2} x^{2}}} d x=\frac{1}{b} \int \frac{1}{\sqrt{\left(a^{2}+t^{2}\right)}} d t$

$=\frac{1}{b} \log \left[t+\sqrt{a^{2}+t^{2}}\right]+c\left\{\right.$ since $\left.\left.\int \frac{1}{\sqrt{\left(a^{2}+x^{2}\right)}} d x=\log \left[x+\sqrt{\left(a^{2}\right.}+x^{2}\right)+c\right\}\right\}$

Put $\mathrm{t}=\mathrm{bx}$

$=\frac{1}{b} \log \left[b x+\sqrt{a^{2}+b^{2} x^{2}}\right]+c$