Factors of 1−x3 are (1−x)(1+x+x2), hence partial fractions of 2x1−x3 are of the form
2x1−x3=A1−x+Bx+C1+x+x2
i.e. 2x1−x3=A(1+x+x2)+(Bx+C)(1−x)1−x3
Comparing coefficients of constant term, x and x2 in numerator
A+C=0, A+B−C=2 and A−B=0
solving these we get A=B=23, C=−23 and hence
2x1−x3=23(1−x)+2x−23(1+x+x2) and
∫2x1−x3dx=∫23(1−x)dx+∫2x−23(1+x+x2)dx
Now ∫23(1−x)dx=23∫11−xdx=−23ln|x−1|
and ∫2x−23(1+x+x2)dx
= ∫2x+13(1+x+x2)dx−∫31+x+x2dx
= 13ln∣∣1+x+x2∣∣−3[2√3tan−1(2x+1√3)]
For former observe ddx(1+x+x2)=2x+1 and for latter, we can have
∫31+x+x2dx=3∫1(x+12)2+34dx=3∫1(x+12)2+(√32)2dx
= 3[2√3tan−1(2x+1√3)]
Hence ∫2x1−x3dx=−23ln|x−1|+13ln∣∣1+x+x2∣∣−3[2√3tan−1(2x+1√3)]
