First note that the denominator of the integrand factorizes as
Thus, we want to use partial fractions to decompose the fraction into the form
![{\displaystyle {\frac {x-8}{x(x-2)^{2}}}={\frac {A}{x}}+{\frac {B}{x-2}}+{\frac {C}{(x-2)^{2}}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/79fda013266082463fa79e613ab51c7b2b99a9f0)
By multiplying through by the denominator of the left hand side and then equating coefficients, we get that
![{\displaystyle A=-2,B=2,C=-3}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/261575c18675dc3237f04f0dcf60021090d19c30)
. Thus we can rewrite our integral as
![{\displaystyle \int _{-3}^{-4}-{\frac {2}{x}}+{\frac {2}{x-2}}-{\frac {3}{(x-2)^{2}}}\,dx.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7409d34122967580f50b1f6a8515b6f608464c53)
Integrating, we get
![{\displaystyle \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).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/0700ebfc59ea9e1c529d6b6a974316987a5840fa)
Answer: