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I've been trying to solve one simple Integral with Euler substitution several times, but can't find where I'm going wrong. The integral is (+ the answer given here, too):

$$\int\frac{1}{x\sqrt{x^2+x+1}} dx=\log(x)-\log(2\sqrt{x^2+x+1}+x+2)+\text{ constant}$$

The problem is, I cannot get this result. Below is my solution of the problem. I've checked it many times, must be something very obvious that I'm missing:

(original image)

Euler Substituion

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

Let $\sqrt{x^2+x+1}=t-x$.

$x^2+x+1=t^2-2xt+x^2$

$x(1+2t)=t^2-1\implies x=\dfrac{t^2-1}{1+2t}$

$dx=\left(\dfrac{t^2-1}{1+2t}\right)'dt=\dfrac{2t(1+2t)-(t^2-1)2}{(1+2t)^2}=\dfrac{2t+4t^2-2t^2+2}{(1+2t)^2}=\dfrac{2(t^2+t+1)}{(1+2t)^2}$

$\sqrt{x^2+x+1}=t-x=t-\dfrac{t^2-1}{1+2t}=\dfrac{t^2+t+1}{1+2t}$

$\implies\displaystyle\int\frac{dx}{x\sqrt{x^2+x+1}}=2\int\frac{\frac{t^2+t+1}{(1+2t)^2}\;dt}{\frac{t^2-1}{1+2t}\cdot\frac{t^2+t+1}{1+2t}}=2\int \frac{1}{t^2-1}\,dt$

$\dfrac{1}{t^2-1}=\dfrac{1}{(t+1)(t-1)}=\dfrac{A}{t+1}+\dfrac{B}{t-1}\implies \begin{eqnarray}&&At-A+Bt+B=1\\&&A+B=0\implies A=-B\\ &&B-A=1\implies B=\frac{1}{2},A=-\frac{1}{2}\end{eqnarray}$

$\implies \displaystyle 2\int \frac{1}{2}\frac{1}{2t-1}\,dt-2\int\frac{1}{2}\frac{1}{t+1}\,dt=\int\frac{1}{t-1}\,dt-\int\frac{1}{t+1}\,dt=$

$=\ln|t-1|-\ln|t+1|=\ln\left|\dfrac{t-1}{t+1}\right|$

$t-x=\sqrt{x^2+x+1}\implies t=\sqrt{x^2+x+1}+x$

$\implies \ln\left|\dfrac{t-1}{t+1}\right|=\ln\left|\dfrac{\sqrt{x^2+x+1}+x-1}{\sqrt{x^2+x+1}+x+1}\right|$

I'll appreciate any help.

Thanks in advance!

  • 1
    I've typed your scanned document in LaTeX to ensure readability, but I may have inadvertently introduced changes from the original. If so, I apologize, and feel free to change any errors I made (or anything else for that matter).2012-06-24
  • 0
    See [here](http://meta.math.stackexchange.com/a/464/264) and [here](http://meta.stackexchange.com/a/70559/161783) for how to format your mathematics expressions with LaTeX, and see [here](http://meta.stackoverflow.com/editing-help) for how to use Markdown formatting. If you need to format more advanced math, there are many excellent LaTeX references on the internet, including Stack Exchange's own [TeX.SE site](http://tex.stackexchange.com/). If you see a piece of LaTeX you want to know the code for on the site, you can right click on it, go to "Show Math As", then choose "TeX Commands".2012-06-24
  • 0
    Hi Zev, thanks for the tip. I'll definitely use LaTeX next time.2012-06-24

3 Answers 3

0

You may use substitution x=1/t, or Euler substitution \sqrt{x^2+x+1}=tx-1

3

Looks about right. In your expression, multiply top and bottom of the thing inside the log by $\sqrt{x^2+x+1}-(x-1)$.

3

What you have done is correct. All you need to do is to rewrite it a different form. $$\begin{align} \ln \left( \sqrt{x^2+x+1} + x - 1\right) & = \ln \left( \dfrac{\left(\sqrt{x^2+x+1} + x - 1 \right) \left(\sqrt{x^2+x+1} - x + 1 \right)}{\left(\sqrt{x^2+x+1} - x + 1 \right)}\right)\\ & = \ln \left(\left(x^2 + x + 1 - (x^2 - 2x + 1) \right) \right) - \ln \left(\sqrt{x^2+x+1} - x + 1 \right)\\ &= \ln(x) - \ln \left(\sqrt{x^2+x+1} - x + 1 \right) \end{align} $$ Hence, $$\begin{align} \ln \left( \dfrac{\sqrt{x^2+x+1} + x - 1}{\sqrt{x^2+x+1} + x + 1}\right) & = \ln \left( \sqrt{x^2+x+1} + x - 1\right) - \ln \left(\sqrt{x^2+x+1} + x + 1 \right)\\ & = \ln(x) - \ln \left(\sqrt{x^2+x+1} - x + 1 \right) -\ln \left(\sqrt{x^2+x+1} + x + 1 \right)\\ & = \ln(x) - \left(\ln \left(\sqrt{x^2+x+1} - x + 1 \right) +\ln \left(\sqrt{x^2+x+1} + x + 1 \right) \right)\\ & = \ln(x) - \ln \left( \left(\sqrt{x^2+x+1}+1 \right)^2 - x^2\right)\\ & = \ln(x) - \ln \left( x^2 + x + 1 + 1 +2 \sqrt{x^2+x+1} - x^2\right)\\ & = \ln(x) - \ln \left( 2 \sqrt{x^2+x+1} + x + 2\right) \end{align}$$

  • 0
    Thanks! I'm wondering, when I have to rewrite the answers - i.e. how do you know the rewriting is enough and the answer is in it's best form.2012-06-24
  • 0
    @cypressx There is no best form. Your answer is equally "best" as wolframalpha.2012-06-24