Why arcsinh$(x)$ is the primitive of $(x^2+1)^{-1/2}$

Question

I have a question: How to calculate the following primitive of $g(x)$. $I=\int g(x)\text{d}x=\int\dfrac{\text{d}x}{\sqrt{x^2+1}}$.

I know that it is equal to the inverse of the sinus hyperbolic function. I want to know the steps to get the function $\sinh^{-1}x$. How to get this result?

Also, I saw in Wikipedia that $\sinh^{-1}x=\log(x+\sqrt{x^2+1})$ but when I calculate $(\sinh^{-1}x)^{\prime}=(\log(x+\sqrt{x^2+1}))^{\prime}$ I do not get $g(x)$. Any explanation please?

Thank you for your help.

Answer 1

You know that the derivative of the inverse function $f^{-1}$ is:

$$\left(f^{-1}\right)'(y)=\frac1{f'(f^{-1}(y))}$$

and recall that $$\cosh^2 y-\sinh^2y=1$$ hence using the last equality we find $$(\sinh^{-1})'(y)=\frac1{\cosh(\sinh^{-1}(y))}=\frac1{\sqrt{1+y^2}}$$

Answer 2

To emphasize how to get there: substitute $$ x = \sinh t, \; \; dx = \cosh t \, dt $$ and $$ \sqrt {1 + x^2} = \sqrt {1 + \sinh^2 t} = \sqrt {\cosh^2 t} = \cosh t. $$ The substituted integral is now $$ \int \frac{1}{\cosh t} \; \cosh t \; dt = \int 1 \, dt = t + C $$ But $x = \sinh t$ and $t = \operatorname{argsinh} x,$ so the integral really is $$ \operatorname{argsinh} x + C $$