Is my integral in fully reduced form? 
I have to integrate this:
  $$\int_0^1 \frac{x-4}{x^2-5x+6}\,dx$$

Now
$$\int_0^1 \frac{x-4}{(x-3)(x-2)}\, dx$$
and by using partial fractions we get
$$\frac{x-4}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$$
$$x-4 = A(x-2) + B(x-3)$$
$$= Ax - 2A + Bx - 3B$$
$$x-4 = (A+B)x - 2A - 3B$$
so
$$A+B=1$$ or $$2X+2B = 2$$
$-B = -2$ and $B = 2$ and $A = -1$
Then
$$\int_0^1 \frac{x-4}{x^2-5x+6} = \int_0^1 \frac{-1}{x-3}dx + \int_0^1 \frac{2}{x-2} dx$$
usub using $u = x-3$ and $du = dx$
so $$ -1 \ln | x-3 | \rbrack_0^1 + 2 \ln |x-2| \rbrack_0^1$$
$$-1 ( \ln 2 - \ln 3) + 2 (\ln 1 - \ln 2)$$
$$-\ln 2 + \ln 3 + 2\ln 1 - 2\ln 2 = -3 * \ln 2 + \ln 3 + 2\ln 1$$
Is there anyway to simplify this further? Wolfram has the answer at $-\ln(8/3)$ and I'm not sure how to simplify to here? How does my work look?

Don't forget that when you add two logarithms with the same base, the expressions under the logarithm sign are multiplied: $\log_b{x}+\log_b{y}=\log_b{(xy)}$. When two logarithms with the same base are subtracted, you divide: $\log_b{x}-\log_b{y}=\log_b{\frac{x}{y}}$. And also don't forget that the power on the expression under the logarithm sign can always be brought in or out of the logaritm: $\log_b{x^a}=a\log_b{x}$.
$$
-( \ln 2 - \ln 3) + 2 (\ln 1 - \ln 2)=-\ln\frac{2}{3}+2\ln\frac{1}{2}=
-\ln\frac{2}{3}+\ln\left(\frac{1}{2}\right)^2=\\
\ln\frac{1}{4}-\ln\frac{2}{3}=
\ln\left(\frac{1}{4}\div\frac{2}{3}\right)=
\ln\frac{3}{8}=
\ln\left(\frac{8}{3}\right)^{-1}=
-\ln\frac{8}{3}.
$$
