Science:Math Exam Resources/Courses/MATH101/April 2017/Question 08/Solution 1

First note that the denominator of the integrand factorizes as $x^{3}-4x^{2}+4x=x(x-2)^{2}.$ Thus, we want to use partial fractions to decompose the fraction into the form

$${\frac {x-8}{x(x-2)^{2}}}={\frac {A}{x}}+{\frac {B}{x-2}}+{\frac {C}{(x-2)^{2}}}.$$

By multiplying through by the denominator of the left hand side and then equating coefficients, we get that $A=-2,B=2,C=-3$. Thus we can rewrite our integral as

$$\int _{-3}^{-4}-{\frac {2}{x}}+{\frac {2}{x-2}}-{\frac {3}{(x-2)^{2}}}\,dx.$$

Integrating, we get

$$\int _{-3}^{-4}-{\frac {2}{x}}+{\frac {2}{x-2}}-{\frac {3}{(x-2)^{2}}}\,dx={\bigg [}-2\ln(x)+2\ln(x-2)+{\frac {3}{x-2}}{\bigg ]}_{-3}^{-4}={\frac {3}{5}}-{\frac {1}{2}}-\ln \left({\frac {100}{81}}\right)={\frac {1}{10}}-\ln \left({\frac {100}{81}}\right).$$

Answer: $\color {blue}{\frac {1}{10}}-\ln \left({\frac {100}{81}}\right)$