1. ## Integration problem

$\displaystyle \int\frac{5x^2}{x^3-1}dx$
Hi I'm trying to integrate the above function. Not sure where to begin, I cant seem to break it up into partial fractions.

2. Rewrite it as $\displaystyle \displaystyle \frac{5}{3}\int{\frac{3x^2}{x^3 - 1}\,dx}$, then make the substitution $\displaystyle \displaystyle u = x^3 - 1$.

3. Originally Posted by flashylightsmeow
$\displaystyle \int\frac{5x^2}{x^3-1}dx$
Hi I'm trying to integrate the above function. Not sure where to begin, I cant seem to break it up into partial fractions.
You don't need partial fractions at all here: observe that $\displaystyle \displaystyle{5x^2=\frac{5}{3}\frac{d(x^3-1)}{dx}$ , so you get an immediate logarithmic integral:

$\displaystyle \displaystyle{\int\frac{5x^2}{x^3-1}\,dx=\frac{5}{3}\int\frac{3x^2}{x^3-1}\,dx$ , and the integral's of the form $\displaystyle \displaystyle{\int\frac{f'}{f}\,dx=\ln|f|+K$ ,

but if you insist in partial fractions then:

$\displaystyle \displaystyle{x^3-1=(x-1)(x^2+x+1)\Longrightarrow \frac{5x^2}{x^3-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}\iff}$

$\displaystyle \displaystyle{\iff 5x^2=A(x^2+x+1)+(Bx+C)(x-1)}$

Now choose nice values of x (say, 0, 1 ,-1) and solve resp. to find the coefficientes A,B,C.

At the end the integral turns out to be pretty easy: it equals $\displaystyle \displaystyle{\frac{5}{3}\ln|x^3-1|+K\,,\,\,K=\mbox{ a constant}}$

Tonio

4. Yes, partial fractions is the hard way- but it can be done. $\displaystyle x^3- 1= (x- 1)(x^2+ x+ 1)$ so that
$\displaystyle \frac{5x^2}{x^3- 1}= \frac{A}{x-1}+ \frac{Bx+ C}{x^2+ x+ 1}$

$\displaystyle 5x^2= A(x^2+ x+ 1)+ (Bx+ C)(x- 1)$

Taking x= 1, 5= 3A so A= 5/3. Taking x= 0, 0= (5/3)(1)+ C(-1) and C= -5/3. Taking x= -1, 5= 5/3+ (-5/3- B)(-2)= 2B+ 15/3= 2B+ 5 so B= 0.

$\displaystyle \frac{5}{3}\int \frac{dx}{x-1}- \frac{5}{3}\int \frac{dx}{x^2+ x+ 1}$

The first is easy to integrate. To integrate the second, complete the square: $\displaystyle x^2+ x+ 1= x^2+ x+ 1/4+ 3/4= (x+1/2)^2+ 3/4$ so the integral is
[tex]-\frac{5}{3}\int \frac{dx}{(x+1/2)^2+ 3/4}[/itex]
and the substitution u= x+ 1/2 makes that
[tex]-\frac{5}{3}\int \frac{du}{u^2+ 3/4}= -\frac{20}{9}\int\frac{du}{(4u^2}/3+ 1}[/itex]
and the final substitution $\displaystyle v= 2u/\sqrt{3}$ makes it
$\displaystyle -\frac{10\sqrt{3}}{9}\int \frac{dv}{v^2+ 1}$

5. Originally Posted by HallsofIvy
Yes, partial fractions is the hard way- but it can be done. $\displaystyle x^3- 1= (x- 1)(x^2+ x+ 1)$ so that
$\displaystyle \frac{5x^2}{x^3- 1}= \frac{A}{x-1}+ \frac{Bx+ C}{x^2+ x+ 1}$

$\displaystyle 5x^2= A(x^2+ x+ 1)+ (Bx+ C)(x- 1)$

Taking x= 1, 5= 3A so A= 5/3. Taking x= 0, 0= (5/3)(1)+ C(-1) and C= -5/3.

Caution: it should be $\displaystyle \displaystyle{C=\frac{5}{3}$

Tonio

Taking x= -1, 5= 5/3+ (-5/3- B)(-2)= 2B+ 15/3= 2B+ 5 so B= 0.

$\displaystyle \frac{5}{3}\int \frac{dx}{x-1}- \frac{5}{3}\int \frac{dx}{x^2+ x+ 1}$
The first is easy to integrate. To integrate the second, complete the square: $\displaystyle x^2+ x+ 1= x^2+ x+ 1/4+ 3/4= (x+1/2)^2+ 3/4$ so the integral is
and the final substitution $\displaystyle v= 2u/\sqrt{3}$ makes it
$\displaystyle -\frac{10\sqrt{3}}{9}\int \frac{dv}{v^2+ 1}$