How to integrate $\int{ dx \over \sqrt{1 + x^2}}$ How to integrate  $dx \over \sqrt{1 + x^2}$?
Answer should be $\log ( x + \sqrt{1 + x^2})$
Please help as possible...
Thank you

Let $\displaystyle 1+x^2=y^2\Rightarrow \sqrt{1+x^2} = y,$ and $2xdx = 2ydy$
$$\displaystyle xdx = ydy\Rightarrow \frac{dx}{y} = \frac{dy}{x} = \frac{d(x+y)}{(x+y)}$$ (Using Ratio and Proportion).
Now $$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx = \int \frac{dx}{y} = \int \frac{d(x+y)}{(x+y)} = \ln \left|x+y\right|+C$$
So $$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx = \ln \left|x+\sqrt{x^2+1}\right|+\mathbb{C}$$
