Write down constants $A$ and $B$ such that for all values of $x$, $x - 1 = A(2x - 4) + B.$ By expressing $x^2 - 4x + 5$ in completed square form, find $\int \frac{x - 1}{x^2 - 4x + 5}\,dx.$ Hence, without using a calculator, show that $\int_{-2}^{2} \frac{|x-1|}{x^2-4x+5}\,dx = \frac{\pi}{2} + \frac{1}{2}\ln\frac{p}{q} - \tan^{-1}\frac{p}{q},$ where $p$ and $q$ are exact real constants to be determined.

1. Expand the right-hand side and compare the coefficients of $x$ to find $A$.

$x - 1 = 2Ax - 4A + B$
By comparing the coefficients of $x$:
$1 = 2A \implies A = \frac{1}{2}$

2. Compare the constant terms to find $B$.

By comparing the constant terms:
$-1 = -4A + B$
Substitute $A = \frac{1}{2}$:
$-1 = -4\left(\frac{1}{2}\right) + B$
$-1 = -2 + B \implies B = 1$

1. Find constants $A$ and $B$ from part (a) and complete the square for the denominator.

From part (a), we have $x - 1 = A(2x - 4) + B$.
Comparing coefficients of $x$: $2A = 1 \implies A = \frac{1}{2}$
Comparing constants: $-4A + B = -1 \implies -2 + B = -1 \implies B = 1$
So, $x - 1 = \frac{1}{2}(2x - 4) + 1$.
Next, completing the square for the denominator:
$x^2 - 4x + 5 = (x - 2)^2 - (-2)^2 + 5 = (x - 2)^2 + 1$

2. Rewrite the integral into two separate standard forms.

Substitute the expressions into the integral:
$\int \frac{x - 1}{x^2 - 4x + 5}\,dx = \int \frac{\frac{1}{2}(2x - 4) + 1}{x^2 - 4x + 5}\,dx$
$= \frac{1}{2} \int \frac{2x - 4}{x^2 - 4x + 5}\,dx + \int \frac{1}{(x - 2)^2 + 1}\,dx$

3. Integrate the first term yielding a natural logarithmic function.

Since the numerator is exactly the derivative of the denominator:
$\frac{1}{2} \int \frac{2x - 4}{x^2 - 4x + 5}\,dx = \frac{1}{2} \ln|x^2 - 4x + 5|$
Since $x^2 - 4x + 5 = (x - 2)^2 + 1 > 0$ for all real $x$, the absolute value signs can be replaced with parentheses:
$= \frac{1}{2} \ln(x^2 - 4x + 5)$

4. Integrate the second term yielding an inverse tangent function and combine for the final answer.

Using the standard result $\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$ with $a=1$ and replacing $x$ with $x-2$:
$\int \frac{1}{(x - 2)^2 + 1}\,dx = \tan^{-1}(x - 2)$
Combining both parts and adding the arbitrary constant $C$:
$\int \frac{x - 1}{x^2 - 4x + 5}\,dx = \frac{1}{2} \ln(x^2 - 4x + 5) + \tan^{-1}(x - 2) + C$

1. Remove the modulus by splitting the range at x = 1

On $[-2,2]$, $|x-1| = 1-x$ for $x\le 1$ and $|x-1| = x-1$ for $x\ge 1$. Hence
$\int_{-2}^{2}\frac{|x-1|}{x^2-4x+5}\,dx=\int_{-2}^{1}\frac{-(x-1)}{x^2-4x+5}\,dx+\int_{1}^{2}\frac{x-1}{x^2-4x+5}\,dx.$

2. Use the antiderivative from part (b)

From (b), $\displaystyle\int\frac{x-1}{x^2-4x+5}\,dx=\tfrac{1}{2}\ln(x^2-4x+5)+\tan^{-1}(x-2)+C$. Let $F(x)=\tfrac{1}{2}\ln(x^2-4x+5)+\tan^{-1}(x-2)$. Then the integral equals
$\big[-F(x)\big]_{-2}^{1}+\big[F(x)\big]_{1}^{2}=F(-2)+F(2)-2F(1).$

3. Evaluate F at the three limits

$F(-2)=\tfrac{1}{2}\ln(4+8+5)+\tan^{-1}(-4)=\tfrac{1}{2}\ln 17-\tan^{-1}4$.
$F(2)=\tfrac{1}{2}\ln(1)+\tan^{-1}(0)=0$.
$F(1)=\tfrac{1}{2}\ln(2)+\tan^{-1}(-1)=\tfrac{1}{2}\ln 2-\tfrac{\pi}{4}$.

4. Combine and identify p and q

$F(-2)+F(2)-2F(1)=\left(\tfrac{1}{2}\ln17-\tan^{-1}4\right)+0-2\left(\tfrac{1}{2}\ln2-\tfrac{\pi}{4}\right)$
$=\tfrac{\pi}{2}+\tfrac{1}{2}\ln17-\ln2-\tan^{-1}4=\tfrac{\pi}{2}+\tfrac{1}{2}\ln\frac{17}{4}-\tan^{-1}4,$
since $\ln2=\tfrac{1}{2}\ln4$. Hence $p=17$, $q=4$.