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$$\int \sqrt{1+x^2}\,dx$$

I've tried some step-by-step calculators, but what they give is way beyond my level. Is there any "simple" way to solve this?

Michael Rozenberg
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username_DNE
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    Integration by parts, the substitution $x=\sinh t$ (or $x=\tan\theta$) or both are the usual ways. And they are simple ways: please show your efforts to let us understand your troubles. – Jack D'Aurizio Oct 14 '17 at 17:09

4 Answers4

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Several people have pointed out that you can write $x=\tan\theta.$ That gets you $dx=\sec^2\theta\,d\theta$ and $\sqrt{1+x^2}=\sec\theta,$ so you have $$ \int\sec^3\theta\,d\theta. $$ This is one of the standard examples of a standard trick: \begin{align} \int \sec^3\theta\,d\theta & = \int(\sec\theta) \Big(\sec^2\theta\,d\theta\Big) = \overbrace{\int u\,dv = uv -\int v\,du}^{\Large\text{integration by parts}} \\[10pt] & = \sec\theta\tan\theta - \int \tan^2\theta\sec\theta\,d\theta \\[10pt] & = \sec\theta\tan\theta - \int(\sec^2\theta-1)\sec\theta\,d\theta \\[10pt] & = \sec\theta\tan\theta - \int\sec^3\theta\,d\theta + \int\sec\theta\,d\theta. \\[15pt] \text{Thus we have } & \int \sec^3\theta\,d\theta = \Big(\text{something}\Big) - \int\sec^3\theta\,d\theta. \\[10pt] \text{Adding the integral to } \\ \qquad \text{both sides, we get } & 2\int\sec^3\theta\,d\theta = \Big(\text{something}\Big). \\[10pt] \text{And so } & \phantom{2}\int\sec^3\theta\,d\theta = \frac 1 2 \Big(\text{something}\Big) \\[10pt] = {} & \frac 1 2 \left( \sec\theta\tan\theta + \int \sec\theta\,d\theta\right). \end{align} Then you have the problem of finding that last integral, which is somewhat challenging as well.

Michael Hardy
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Hint: try substitute $$x=\tan { \theta } $$

haqnatural
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$x=\frac{e^t-e^{-t}}{2}$ also helps.

Michael Rozenberg
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  • ...which is just another way to write $\sinh(t)$, though I do give you credit for that since given that the OP did not understand the solution given by the step-by-step calculators, he/she may not have heard of hyperbolic trig functions. – projectilemotion Oct 14 '17 at 17:56
  • @projectilemotion I think this way more easier than a way with $x=\tan{t}$ because there we get $\sqrt{\frac{1}{\cos^2t}}=\frac{1}{|\cos{t}|}$, which is not so easy. – Michael Rozenberg Oct 14 '17 at 18:04
  • the OP has indeed not heard/learned of hyperbolic trig functions; I'll try this out. Thanks! – username_DNE Oct 14 '17 at 19:14
  • @username_DNE You are welcome! Good luck! – Michael Rozenberg Oct 14 '17 at 19:16
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This is just straight up trigonometric substitution with $x=\tan(\theta)$ and $dx=\sec^{2}(\theta)\:d\theta$. Then use the identity $1+\tan^{2}(\theta)=\sec^{2}(\theta)$ to eliminate the square root and integrate from there.

Sir_Math_Cat
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