Answer
$a)\int\frac{dx}{\sqrt{x^2+a^2}} = \ln(x+\sqrt{x^2+a^2})+ C $
$b)\int\frac{dx}{\sqrt{x^2+a^2}} = arcsinh \left ( \frac{x}{a} \right )+ C $
Work Step by Step
$ a) $
$I=\int\frac{dx}{\sqrt{x^2+a^2}}$
$\begin{Bmatrix}
x=a \tan\theta & x^2=a^2 \tan^2\theta \\
\frac{dx}{d\theta}=a\sec^2\theta & dx=a\sec^2\theta {d\theta}
\end{Bmatrix}$
Integration by substitution
$=\int\frac{a\sec^2\theta {d\theta}}{\sqrt{a^2
\tan^2\theta+a^2}}$
$=\int\frac{a\sec^2\theta {d\theta}}{\sqrt{a^2( \tan^2\theta+1)}}$
$=\int\frac{a\sec^2\theta {d\theta}}{\sqrt{a^2 \sec^2 }}$
$=\int\frac{a\sec^2\theta {d\theta}}{a\sec }$
$=\int{\sec {d\theta}}$
$=\ln(\sec\theta+\tan\theta)+C1$
$\sec\theta=\frac{{\sqrt{x^2+a^2}}}{a}$
$\tan\theta=\frac{x}{a}$
$=\ln(\frac{{\sqrt{x^2+a^2}}}{a}+\frac{x}{a})+ C1$
$=\ln({\sqrt{x^2+a^2}}+{x})-\ln(a)+ C1$
$C=-\ln(a)+C1$
$I=\ln({\sqrt{x^2+a^2}}+{x})+ C1$
$ b) $
$I=\int\frac{dx}{\sqrt{x^2+a^2}}$
$\begin{Bmatrix}
x=a \sinh t & x^2=a^2 \sinh^2t \\
\frac{dx}{dt}=a\cosh t & dx=a\cosh t {dt}
\end{Bmatrix}$
$=\int\frac{a\cosh t {dt}}{\sqrt{a^2
\sinh^2t+a^2}}$
$=\int\frac{a\cosh t {dt}}{\sqrt{a^2( sinh^2t+1)}}$
$=\int\frac{a\cosh t {dt}}{\sqrt{a^2 \cosh^2 }}$
$=\int\frac{a\cosh{dt}}{a\cosh }$
$=\int{1{dt}}$
$=t+C$
$t=arcsinh \left ( \frac{x}{a} \right )$
$=arcsinh \left ( \frac{x}{a} \right )+C$
