(8x-5)/(x^3-1) = (8x-5)/((x-1)(x^2+x+1))
= A/(x-1) + (Bx+C)/(x^2+x+1)
follow the normal partial fraction solution to find that
∫((8x-5)/(x^3 -1))dx = ∫(1/(x-1) - (x-6)/(x^2+x+1) dx
Now, if u = x^2+x+1, du=2x+1 dx
x-6 = 1/2 (2x-1) - 11/2
So now we have to find
∫ 1/(x-1) dx = ln(x-1)
∫ 1/2 (du/u) = 1/2 ln(x^2+2x+1)
Now, just one step to go
x^2+x+1 = (x + 1/2)^2 + 3/4
You know that ∫1/(a^2+u^2) = arctan(u/a)
So the final result is
ln(1-x) - 1/2 ln(x^2+x+1) + 13/√3 arctan (2x+1)/√3 + C
How do I decompose and solve this integral?
∫((8x-5)/(x^3 -1))dx
2 answers
Thank oobleck