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.
 A: Note that $$\frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}=\frac{3x^3 - 2x^2 +3x -2+\color{red}{5x^2-8x+6}}{3x^3 - 2x^2 +3x -2}=1+\frac{5x^2-8x+6}{3x^3 - 2x^2 +3x -2}$$ and $$3x^3-2x^2+3x-2=(3x-2)(x^2+1)$$ so $$\int_1^2\frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\,dx=[x]_1^2+\int_1^2\frac{5x^2-8x+6}{(3x-2)(x^2+1)}\,dx$$ Now can you use partial fractions?
A: \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 $x = 2/3$
$$\frac{\frac{26}{9}}{\frac{13}{9}} = 2 = A \implies A = 2 $$
$$\frac {3x^3 + 3x^2 - 5x + 4}{3x^3 -2x^2 + 3x - 2} = 1 + \frac {2}{3x- 2} + \frac {x-2}{x^2+1}$$
A: Apologies if it's unclear - I'm not sure how to get the polynomial division in Mathjax, if someone could advise me on how to do this it would be appreciated.

From here I'll use Mathjax however:
$$5x^2-8x+6=A(x^2+1)+B(3x-2)$$
Let $A=px+q, B=rx+s$. If $p>0$, this equation transforms into a cubic so $p=0$.
Hence $qx^2+q+(rx+s)(3x-2)=5x^2-8x+6$
Expanding and collecting like terms we get:
$$q+3r=5$$
$$3s-2r=-8$$
$$q-2s=6$$
I instantly checked $q=2$ and $r=1$ (seemed like an obvious first choice given the first statement) and it works with $s=-2$. 
Hence: $$\frac{5x^2-8x+6}{(x^2+1)(3x-2)}=1+\frac{2}{3x-2}+\frac{x-2}{x^2+1}$$
Integrate these seperately and sum the results:
$$\int_1^2{1 dx}=\bigg[x\bigg]^2_1=1$$
$$\int_1^2{\frac{2}{3x-2}dx}=2\int_1^2{\frac{1}{3x-2}dx}=2\bigg[\frac 13\ln|3x-2|\bigg]^2_1=\frac 23\ln(4)=\frac 43\ln(2)$$
$$\int_1^2{\frac{x-2}{x^2+1}dx}=\int_1^2{\frac{x}{x^2+1}dx}-\int_1^2{\frac{2}{x^2+1}dx}$$
Note the first of those is of the form $\int{\frac{f'(x)}{2f(x)}dx}$, this integral therefore is $\bigg[\frac 12\ln|x^2+1|\bigg]^2_1=\frac 12(\ln5-\ln2)$
The second half requires $u$-substitution. Let $x=\tan(u)$, then $dx=\sec^2(u) du$. Our integral becomes:
$$2\int_{\tan^{-1}(1)}^{\tan^{-1}(2)}{\frac{1}{sec^2(u)}\cdot \sec^2(u) du}=2\bigg[1\bigg]^{\tan^{-1}(2)}_{\tan^{-1}(1)} \because 1+\tan^2(u)=\sec^2(u)$$
$$=2\tan^{-1}(2)-\frac \pi2\approx 0.6435 \text{(this is also $\tan^{-1}\bigg(\frac 34\bigg)$)}$$
Summing all we get:$$1+\frac 56 \ln(2)+ \frac 12 \ln(5)-2\tan^{-1}(2)+\frac \pi2$$
