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?
 A: It is fine, except that I don't see where that $3$ at the last line comes from. Anyway,\begin{align}-\bigl(\log(2)-\log(3)\bigr)+2\bigl(\log(1)-\log(2)\bigr)&=\log(3)-3\log(2)\\&=\log(3)-\log(2^3)\\&=\log\left(\frac38\right)\\&=-\log\left(\frac83\right).\end{align}
A: Use $ln(a) + ln(b) = ln(a\cdot b)$ and $aln(b) = ln(b^a)$
You get $ln(\frac{1}{2} \cdot \frac{3}{1} \cdot \frac{1^2}{1} \cdot \frac{1}{2^2})$
A: 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: $\log_b{x^a}=a\log_b{x}$.
$$
-( \ln 2 - \ln 3) + 2 (\ln 1 - \ln 2)=-\ln\frac{2}{3}+2\ln\frac{1}{2}=
-\ln\frac{2}{3}+\ln\left(\frac{1}{2}\right)^2=\\
\ln\frac{1}{4}-\ln\frac{2}{3}=
\ln\left(\frac{1}{4}\div\frac{2}{3}\right)=
\ln\frac{3}{8}=
\ln\left(\frac{8}{3}\right)^{-1}=
-\ln\frac{8}{3}.
$$
A: First, $\log 1= 0$. And there is a mistake when you compute the definite integral:
$$
-\log 2 +\log 3 -2\log2 = -3\log 2 +\log 3 = -\log 8+\log3 = -\log\frac 83.
$$
