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.
Your integral is
$\displaystyle \frac{5}{3}\int \frac{dx}{x-1}- \frac{5}{3}\int \frac{dx}{x^2+ x+ 1}$
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
$\displaystyle -\frac{10\sqrt{3}}{9}\int \frac{dv}{v^2+ 1}$