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I got the integral $$\int \frac{6x-1}{3x^3-4} \, dx$$ as an exercise for exam preparation. I tried to get closer to the result with third binomial formula, but could not get anywhere. So I tried to find the solution with wolframalpha, but this one seems overly complex. Is there a simple solution for this integral?

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  • $\begingroup$ Did you check WA's solution is correct by taking the derivative? If you did so then the only hope you have is to make some algebraic manipulations to put the solution in a more "nice" form, yet it will likely remain a nasty solution... $\endgroup$ – DonAntonio Jan 2 '13 at 12:25
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$$\int \frac{6x-1}{3x^3-4} \, dx$$

$$\frac{1}{4}\int \frac{6x-1}{\frac{3}{4}x^3-1} \, dx$$

$\frac{3}{4}x^3=u^3$

$\sqrt[3]{\frac{3}{4}}x=u$

$\sqrt[3]{\frac{4}{3}}u=x$

$\sqrt[3]{\frac{4}{3}}du=dx$

$$\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{6\sqrt[3]{\frac{4}{3}}u-1}{u^3-1} \, du=\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{A}{u-1} +\frac{B}{u-\epsilon}+\frac{C}{u-\epsilon^2}\, du$$

where $\epsilon=e^{\frac{2\pi i}{3} }=-\frac{1}{2}+i\frac{\sqrt3}{2}$

$\epsilon^2=e^{\frac{4\pi i}{3} }=-\frac{1}{2}-i\frac{\sqrt3}{2}$

$$\frac{6\sqrt[3]{\frac{4}{3}}u-1}{u^3-1}=\frac{A}{u-1} +\frac{B}{u-\epsilon}+\frac{C}{u-\epsilon^2}$$

$f(u)=6\sqrt[3]{\frac{4}{3}}u-1$

$A=f(1)=6\sqrt[3]{\frac{4}{3}}-1$

$B=f(\epsilon)=6\epsilon\sqrt[3]{\frac{4}{3}}-1$

$C=f(\epsilon^2)=6\epsilon^2\sqrt[3]{\frac{4}{3}}-1$

$$\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{A}{u-1} +\frac{B}{u-\epsilon}+\frac{C}{u-\epsilon^2}\, du=$$ $$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-1} \, du+\frac{B\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon}\, du+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon^2}\, du=$$

$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-1} \, du+\frac{B\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon}\, du+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon^2}\, du=$$

$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\ln {(u-1)} +\frac{B\sqrt[3]{\frac{4}{3}}}{4}\ln {(u-\epsilon)}+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\ln {(u-\epsilon^2)}+c=$$

$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\ln {(\sqrt[3]{\frac{3}{4}}x-1)} +\frac{B\sqrt[3]{\frac{4}{3}}}{4}\ln {(\sqrt[3]{\frac{3}{4}}x-\epsilon)}+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\ln {(\sqrt[3]{\frac{3}{4}}x-\epsilon^2)}+c=$$

After that you will need to do some calculations and also need to do transform of the complex value of $\ln$ couple to $arctan$

Use the formula for that transform $\arctan x = \frac{1}{2}i\left(\ln\left(1-i\,x\right)-\ln\left(1+i\,x\right)\right) $

Ref :http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

The most easiest way for me is that way. Maybe someone else can offer quicker way.

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Use partial fractions. Factor the denominator as $$3x^3-4=3(x^3-4/3)=3(x-(4/3)^{1/3})(x^2+(4/3)^{1/3}x+(4/3)^{2/3}),$$ then set up with $A$ over the linear factor, and $Bx+C$ over the quadratic factor.

The answer is messy, so the values of $A,B,C$ will be complicated. However the integral will be doable using $\ln(u),\arctan(u)$ as usual with partial fractions technique.

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