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.


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


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

  • $\begingroup$ Thank you very much. Why when I use the formula with logarithm I cannot get the right answer? $\endgroup$ – zighalo Apr 4 '14 at 16:24
  • 1
    $\begingroup$ If you derivate the $\log$ expression you find $\frac1{\sqrt{x^2+1}}$ so what you can deduce? $\endgroup$ – user63181 Apr 4 '14 at 16:27
  • $\begingroup$ I deduce that I have the right answer. But when I derivate the $\log$ expression I get: $\dfrac{1+(1/2)2x(x^2+1)^{-1/2}}{x+\sqrt{x^2+1}}$. I tried to simplfy it to get $\dfrac{1}{\sqrt{x^2+1}}$ but I cannot. Where is my wrong move? $\endgroup$ – zighalo Apr 4 '14 at 16:31
  • $\begingroup$ Your calculus is true just multiply numerator and denominator by $(x^2+1)^{1/2}$ and simplify the numerator. $\endgroup$ – user63181 Apr 4 '14 at 16:34
  • 1
    $\begingroup$ Ohh I got it. Thank you very mcuh for your help. $\endgroup$ – zighalo Apr 4 '14 at 16:46

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 $$ \operatorname{argsinh} x + C $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.