It is not difficult to verify that $$ \frac{\mathrm d}{\mathrm dx} \left[ \log\Big(x+\sqrt{x^2+1}\Big) \right] = \frac{1}{\sqrt{1+x^2}} $$ for $x\geq 0$, say.

How would one calculate the indefinite integral $$ \int \frac{1}{\sqrt{1+x^2}} \ \!\mathrm dx$$ without knowing this? I have tried many of the usual tricks, without success.

The title of the question is chosen because Mathematica outputs $\text{Arcsinh}(x)+C$ as the answer.


You can do a substitution for $x=\tan\theta$ and $\mathrm dx=\sec^2\theta\,\mathrm d\theta$ to get

$$ \int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm d\theta. $$

Then use that $1+\tan^2\theta=\sec^2\theta$ to get the integral

$$ \int\sec\theta\,\mathrm d\theta. $$

EDIT: In response to the comment by Sasha: This is one of the standard integrals in a calculus class. I usually just derive it in class and have students memorize it along with the other trigonometric functions. Just multiply top and bottom by $\sec\theta + \tan\theta$ to get

$$ \int \frac{\sec^2\theta + \tan\theta\sec\theta}{\sec\theta+\tan\theta}\mathrm d\theta. $$

Then $u=\sec\theta + \tan\theta$, $\mathrm du=\sec\theta\tan\theta + \sec^2\theta\;\mathrm d\theta.$ The answer is then $\ln|\sec\theta+\tan\theta|+C$. Putting this in terms of $x$ one gets $\ln|\sqrt{1+x^2}+x|+C$.


One would use Euler's substitution. Also on PlanetMath.

Alternatively, one could use $x = \sinh(t)$, because $1+x^2 = 1+\sinh^2(t) = \cosh^2(t)$, and because $\mathrm{d}x = \sinh^\prime(t) \mathrm{d}t = \cosh(t) \mathrm{d}t$. Therefore, using that $\cosh(t) >0 $ for real $t$:

$$ \int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \int \frac{\cosh(t)}{\sqrt{\cosh^2(t)}} \mathrm{d} t = \int \mathrm{d} t = t + C $$ Substituting back $\int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \operatorname{arcsinh}(x) + C$

  • $\begingroup$ Your answer gives arcsinh($x$), not the function involving the logarithm. I guess after you'd have to further solve $e^z-e^{-z} = 2w$ which ends up being a quadratic in $e^z$ to get the answer I want. $\endgroup$ – Jim Oct 7 '11 at 21:07
  • $\begingroup$ @Jum The answer you want is provided by Euler's substitution. I was explaining how Mathematica arrived at arcsine hyperbolic. $\endgroup$ – Sasha Oct 7 '11 at 21:35

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.