I have found other variations of the above integral such as:
$\int\frac{dx}{\sqrt{1+x^2}}=\int{\sec{\theta}d{\theta}}=\log{|\sec{\theta}+\tan{\theta}|}+c$ ,using appropriate substitution
Kindly let know what substitution can be made to get the result in title.
On
We have two answers by two substitutions:
Putting $x=\tan \theta$ gives $$ \begin{aligned} I =\ln |\sec \theta+\tan \theta|+C_1=\ln \left(\sqrt{1+x^2}+x\right)+C_1 \end{aligned} $$ Putting $x=\sinh \theta$ gives
$$I=\operatorname{arcsinh} x+C_2$$ Actually, we can prove that the answers by letting $$y= \ln \left(\sqrt{1+x^2}+x\right) $$
By the definition of logarithm, we have $$ \begin{aligned} & e^y=\sqrt{1+x^2}+x \textrm{ and } \\ & e^{-y}=\frac{1}{\sqrt{1+x^2}+x} \cdot \frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}-x}= \sqrt{1+x^2}-x \end{aligned} $$ Adding them together yields $$ \sinh y=\frac{e^y-e^{-y}}{2}=x \Rightarrow \ln \left(\sqrt{1+x^2}+x\right) =y=\operatorname{arcsinh} x $$
Let $$y=\sinh^{-1}x$$ $$\implies x=\sinh(y)$$ $$\implies \frac{dx}{dy}=\cosh(y)$$ $$\implies \frac{dy}{dx}=\frac{1}{ \cosh(y)}$$ But we also know that $$\cosh(y)=\sqrt{1+\sinh^2(y)}=\sqrt{1+x^2}$$ So $$\frac{d}{dx}\sinh^{-1}x=\frac{1}{\sqrt{1+x^2}}$$ So integrating both sides we get $$\int\frac{dx}{\sqrt{1+x^2}}=\sinh^{-1}(x)+c $$