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Explanation:

Step1: Factor the denominator

First, factor $x^{2}-7x + 10=(x - 2)(x - 5)$. Then, use partial - fraction decomposition. Let $\frac{x - 8}{x^{2}-7x + 10}=\frac{x - 8}{(x - 2)(x - 5)}=\frac{A}{x - 2}+\frac{B}{x - 5}$. Cross - multiply: $x - 8=A(x - 5)+B(x - 2)$. Set $x = 2$: $2-8=A(2 - 5)+B(2 - 2)$, so $-6=-3A$, and $A = 2$. Set $x = 5$: $5-8=A(5 - 5)+B(5 - 2)$, so $-3 = 3B$, and $B=-1$. Then $\frac{x - 8}{x^{2}-7x + 10}=\frac{2}{x - 2}-\frac{1}{x - 5}$.

Step2: Integrate term - by - term

$\int_{0}^{1}\frac{x - 8}{x^{2}-7x + 10}dx=\int_{0}^{1}(\frac{2}{x - 2}-\frac{1}{x - 5})dx$. Using the integral formula $\int\frac{1}{u}du=\ln|u|+C$, we have $\int_{0}^{1}(\frac{2}{x - 2}-\frac{1}{x - 5})dx=2\int_{0}^{1}\frac{1}{x - 2}dx-\int_{0}^{1}\frac{1}{x - 5}dx$. $2\int_{0}^{1}\frac{1}{x - 2}dx=2[\ln|x - 2|]{0}^{1}=2(\ln|1 - 2|-\ln|0 - 2|)=2(\ln1-\ln2)=-2\ln2$. $-\int{0}^{1}\frac{1}{x - 5}dx=-[\ln|x - 5|]_{0}^{1}=-(\ln|1 - 5|-\ln|0 - 5|)=-(\ln4-\ln5)=\ln5-\ln4$.

Step3: Combine the results

$-2\ln2+\ln5-\ln4=-2\ln2+\ln5 - 2\ln2=\ln5-4\ln2=\ln5-\ln16=\ln\frac{5}{16}$.

Answer:

$\ln\frac{5}{16}$