<s>[INST] Integration of $ \int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2} \mathrm dx$ Evaluate $$\int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\mathrm  dx$$I believe I should use the partial fractions technique to evaluate this integral, but I am not getting anywhere when I try it.
 [/INST] \begin{align}  \frac {3x^3 + 3x^2 - 5x + 4}{3x^3 -2x^2 + 3x - 2}
& = \frac {(3x^3 - 2x^2 + 3x - 2) + 5x^2 - 8x + 6}{3x^3 -2x^2 + 3x - 2}\\ \\
&\ = 1 + \frac{5x^2 - 8x + 6}{3x^3 -2x^2 + 3x - 2} \end{align}
Factor the denominator
$$3x^3 -2x^2 + 3x - 2 = (3x-2)(x^2 + 1)$$
$$\frac {5x^3 -8x^2 +6}{(3x-2)(x^2+1)} = \frac {A}{3x-2}+ \frac {Bx+C}{x^2 + 1}$$
The "obvious" way to do this..
$$5x^2 -8x +6 = Ax^2 + A + 3Bx^2 + (-2B+3C)x - 2C$$
Giving a system of equations:
\begin{align} A+3B &= 5\\
-2B+3C &= -8\\
A - 2C &= 6 \end{align}
Here is a trick which you may, or may not find easier.
$$\frac {5x^3 -8x^2 +6}{(3x-2)(x^2+1)} = \frac {A}{3x-2}+ \frac {Bx+C}{x^2 + 1}$$
Multiply through by $(3x-2)$
$$\frac {5x^3 -8x^2 +6}{(x^2+1)} = A+ \frac {Bx+C}{x^2 + 1}(3x-2)$$
and evaluate at $x = 2/3$
$$\frac{\frac{26}{9}}{\frac{13}{9}} = 2 = A \implies A = 2 $$
$$\frac {3x^3 + 3x^2 - 5x + 4}{3x^3 -2x^2 + 3x - 2} = 1 + \frac {2}{3x- 2} + \frac {x-2}{x^2+1}$$
</s>