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I'm trying to solve this integral:

$$\int \frac{x}{x^3-1}\,\mathrm dx$$

What I did was: $$\int \frac{x}{(x-1)(x^2+x+1)}\,\mathrm dx.$$

$$\frac{x}{(x-1)(x^2+x+1)} = \frac{a}{x-1}+ \frac{bx+c}{x^2+x+1}$$

Then I got this in the numerator:

$$ax^2+ax+a+bx^2-bx+cx-c $$

$$a+b=0;a-b+c=1; a-c=0 $$

$$a=c=\frac{1}3 \qquad b=-\frac{1}3$$

Then I wrote: $$\frac{1}3\int \frac{1}{x-1}\,\mathrm dx-\frac{1}3\int\frac{x-1}{x^2+x+1} \, \mathrm dx$$

so the first one is just $\frac{1}{3}\ln|x-1|$. Which makes my calculations already wrong, most likely.

With the second one I tried a few different things ( involving u-substitution mostly) and got stuck.

I know I'm supposed to get this: $$\frac{1}6\ln \frac{(x-1)^3}{x^2+x+1}+\frac{1}{\sqrt{3}} \arctan\frac{2x+1}{\sqrt{3}}$$

What have I already done wrong? What am I supposed to do?

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Hint:

$$\int\dfrac{x - 1}{x^2 + x + 1}\,\mathrm dx\equiv\int\dfrac{2x + 1}{2(x^2 + x + 1)} - \dfrac{3}{2(x^2 + x + 1)}\,\mathrm dx$$

Then, let $u = x^2 + x + 1\implies\mathrm du = 2x + 1\,\mathrm dx$. So,

$$\int\dfrac{2x + 1}{x^2 + x+ 1}\,\mathrm dx\equiv\int\dfrac1u\,\mathrm du.$$

Notice that $$\int\dfrac1{x^2 + x + 1}\,\mathrm dx\equiv\int\dfrac1{\left(x + \frac12\right)^2 + \frac34}\,\mathrm dx$$ Let $v = \dfrac{2x + 1}{\sqrt3}\implies\mathrm dx=\dfrac{\sqrt3}2\,\mathrm dv$. So,

$$\int\dfrac1{\left(x + \frac12\right)^2 + \frac34}\,\mathrm dx\equiv\dfrac2{\sqrt3}\int\dfrac1{v^2 + 1}\,\mathrm dv.$$

Edit: $$\begin{align}\left(x + \dfrac12\right)^2 + \dfrac34 = \left(\dfrac{2x + 1}2\right)^2 + \dfrac34 &= \dfrac14\left(3\left(\dfrac{2x + 1}{\sqrt 3}\right)^2 + 3\right) \\ &= \dfrac34\left(\left(\dfrac{2x + 1}{\sqrt 3}\right)^2 + 1\right)\end{align}$$

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    $\begingroup$ How am I supposed to figure out that my v should be 2x+1/sqrt3 ? where does that square root come from? $\endgroup$ – Nullius Feb 19 at 21:17
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    $\begingroup$ @Nullius see the edit. Do you understand now? $\endgroup$ – an4s Feb 19 at 21:44
  • $\begingroup$ Is it because we have 3/4 that we want to get rid of to get the v^2+1? Never would've have gotten it without this explanation, been thinking about maybe some rules that I might not know instead of just this. Thank you! $\endgroup$ – Nullius Feb 19 at 21:57
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    $\begingroup$ Yes, we want to make it so that we can get a form that is easily integratable. $\endgroup$ – an4s Feb 19 at 23:20
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The second integral can be carried as follows,

$$\int\frac{1-x}{x^2+x+1}dx=\int\frac{-(x+\frac12)+\frac32}{(x+\frac12)^2+\frac34}dx$$ $$=-\frac12\int\frac{d[(x+\frac12)^2]}{(x+\frac12)^2+\frac34} +\frac32\int\frac{dx}{(x+\frac12)^2+\frac34}$$ $$=-\frac12\ln\left[(x+\frac12)^2+\frac34\right]+\frac{1}{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}$$

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You did right, but gave up soon. For $$\int\frac{x-1}{x^2+x+1}$$ write $$\frac{x-1}{x^2+x+1}=\frac{1}{2}\frac{2x-2}{x^2+x+1}=\frac{1}{2}\frac{2x+1}{x^2+x+1} +\frac{1}{2}\frac{-3}{x^2+x+1}$$ then the integral becomes $$\int\frac{x-1}{x^2+x+1}dx=\frac{1}{2}\ln(x^2+x+1)+\frac{3}{2}\int\frac{dx}{x^2+x+1}$$ and then...

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  • $\begingroup$ This might be a small thing, but the fact that you wrote 2x-2 first (explaining the 2(x-1)/2 and how we get to 2x+1 thing is really really helpful, I was stuck with that for some reason. $\endgroup$ – Nullius Feb 19 at 20:42
  • $\begingroup$ @Nullius So, would you press a +1 :) $\endgroup$ – Qurultay Feb 19 at 20:45
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You can write that $$ x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=\frac{3}{4}\left(1+\left(\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right)^2\right) $$

and you can then put your fraction under the form $$ \frac{u'}{1+u^2} $$ that integrated to an arctan.

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