How do I integrate this?

Question

(1-2x) / (2+3x+3x^2) dx

Answer 1

First separate into elementary integrable expression:
That is here to find a expression where the nominator is the derivative of the denominator plus a constant divided by the denominator.
∫ (1-2x)/(3x²+3x+2) dx
= -1/3 · ∫ (2x+1-2)/(x²+x+2/3) dx
= -1/3 · ∫ (2x+1)/(x²+x+2/3) dx + 2/3 · ∫ 1/(x²+x+2/3) dx

For the first integral the rule
∫ f'(x)/f(x) dx = ln (f(x)) applies. Hence
-1/3 · ∫ (2x+1)/(x²+x+2/3) dx = -1/3 ·ln(x²+x+2/3)

The second integral could be transformed in an integral of the type:
∫ 1/ (x²+1) dx = arctan(x)
Here
2/3 · ∫ 1/(x²+x+2/3) dx
= 2/3 · ∫ 1/(x²+x+¼-¼+2/3) dx
= 2/3 · ∫ 1/(x²+x+1/4+5/12) dx
= 2/3 · ∫ 1/((x+½)² +5/12) dx
= 2/3 ·12/5 · ∫ 1/([√(12/5) · (x+½)]² + 1) dx
with the substitution u = √(12/5) · (x+½) => dx=1/√(12/5) dx
= 2/3 ·√(12/5) · ∫ 1/ (u²+1) du
= 2/3 ·√(12/5) · ∫ 1/ (u²+1) du
= 4/√15 · ∫ 1/ (u²+1) du
= 4/√15 · arctan(u)
= 4/√15 · arctan(√(12/5) · (x+½) )

Therefore the result is:
∫ (1-2x)/(3x²+3x+2) dx = -1/3 ·ln(x²+x+2/3) + 4/√15 · arctan(√(12/5) · (x+½) ) + C

Answer 2

Let
u=2+3x+3x^2
du=3(1+2x)

the numerator must be 1+2x
then
integral of du/3u

=1/3 ln |2+3x+3x^2|+C