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

I tried to use trigonometric substitution, let $x=\tan^2 U$,but it is difficult to proceed, can anyone help me how to integrate this indefinite integral?

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  • $\begingroup$ Looks like $(x+\frac 12)+\frac 34$ for the inner expression. Perhaps $\tan u$ is still appropriate? $\endgroup$ – abiessu Jan 5 '15 at 1:25
  • $\begingroup$ @abiessu: I think you meant to square those parentheses. $\endgroup$ – Rory Daulton Jan 5 '15 at 1:41
  • $\begingroup$ @RoryDaulton: I did indeed. That's one weakness I see in this version of the SE app is that comments don't get MathJAX interpolation. $\endgroup$ – abiessu Jan 5 '15 at 3:29
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Hint: 1) Complete the square in the bottom, writing the part under the radical as $(x+a)^2+b$.

2) Use substitution rule with $u=x+a$.

3) Now you have a form where a good trigonometric substitution will work well.

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  • $\begingroup$ thank you very much ,I got it ! $\endgroup$ – user143997 Jan 5 '15 at 1:30
  • $\begingroup$ I'm afraid it will be very complicated with a trigonometric substitution. What about a hyperbolic substitution? $\endgroup$ – Bernard Jan 5 '15 at 1:48
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In fact that's a very typical variation of $$\int \frac{dx}{\sqrt{1+x^2}}=arcsh(x)+C$$ where$$arcsh(x)=ln(x+\sqrt{1+x^2})$$ is the inverse function of $$sh(x)=\frac{e^x-e^{-x}}{2}$$ With the above pattern, all you have to do is to let $u=x+\frac{1}{2}$,and then everything is so natural.

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  • $\begingroup$ That function is usually called $\sinh(x)$, I believe. $\endgroup$ – Akiva Weinberger Jan 5 '15 at 2:06
  • $\begingroup$ Maybe. But I have been accustomed to using this simpler form. :D $\endgroup$ – Vim Jan 5 '15 at 2:08
  • $\begingroup$ I think $\sinh$ and $\cosh$ make more sense, to highlight the similarities between them and $\sin$ and $\cos$. Such as: $(\cos x\pm i\sin x)^a=\cos ax\pm i\sin ax$, and $(\cosh x\pm\sinh x)^a=\cosh ax\pm\sinh ax$. $\endgroup$ – Akiva Weinberger Jan 5 '15 at 2:11
  • $\begingroup$ @columbus8myhw The form $\text{sh}\,x$ was used in the Soviet Union (and is being used today in Russia). $\endgroup$ – Artem Jan 5 '15 at 2:12
  • $\begingroup$ @Artem Interesting. I did not know that! $\endgroup$ – Akiva Weinberger Jan 5 '15 at 2:14
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You write the radicand in canonical form: $$ x^2+x+1=\Bigl(x+\dfrac12\Bigr)^2+\dfrac 3 4 =\dfrac 3 4\biggl(\biggl(\dfrac{2x+1}{\sqrt 3}\biggr)^2+1\biggr) $$ Then the substitution: $\,\,u=\dfrac{2x+1}{\sqrt 3}$ results in $$\DeclareMathOperator\arsinh{arsinh} \int\dfrac{\mathrm d\,u}{\sqrt{u^2+1}}= \arsinh u = \ln\Bigl(u+\sqrt{u^2+1}\Bigr).$$

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  • $\begingroup$ It seems that there are some problems with constants in the calculations. Anyway, the easiest way to compute this integral without using hyperbolic or trigonometric substitutions is to put $t-x=\sqrt{x^2+x+1}$. $\endgroup$ – Artem Jan 5 '15 at 1:55
  • $\begingroup$ ?? I'm afraid I don't follow you. $\endgroup$ – Bernard Jan 5 '15 at 2:00
  • $\begingroup$ What is so difficult to follow? 1. Your computations contain mistakes. 2. Your approach works if one knows that $(\sinh^{-1} x)'=\frac{1}{\sqrt{1+x^2}}$. Otherwise, the best way to proceed as I indicated. $\endgroup$ – Artem Jan 5 '15 at 2:04
  • $\begingroup$ Thanks for pointing the mistake. For point 2), it's standard knowledge about hyperbolic functions and their inverses. But I still don't see how you make your substitution, sorry… $\endgroup$ – Bernard Jan 5 '15 at 2:13
  • $\begingroup$ For point 2 - it is not, actually. It is the ridiculous way to teach calculus by memorizing an incredible number of unnecessary formulas. For my substitution you need to square both sides, cancel what you can cancel, express $x$ as a function of $t$, find $dt$ and $t-x$, and plug everything in it. If you've never seen it, try it, you will see how easy the result follows (it is often called Euler's substitution) $\endgroup$ – Artem Jan 5 '15 at 2:16

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