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

First note that the denominator of the integrand factorizes as $x^{3} - 4 x^{2} + 4 x = x (x - 2)^{2} .$ 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 $\int_{- 3}^{- 4} - \frac{2}{x} + \frac{2}{x - 2} - \frac{3}{(x - 2)^{2}} d x .$

Integrating, we get $\int_{- 3}^{- 4} - \frac{2}{x} + \frac{2}{x - 2} - \frac{3}{(x - 2)^{2}} d x = [- 2 \ln (x) + 2 \ln (x - 2) + \frac{3}{x - 2}]_{- 3}^{- 4} = \frac{3}{5} - \frac{1}{2} - \ln (\frac{100}{81}) = \frac{1}{10} - \ln (\frac{100}{81}) .$

Answer: $\frac{1}{10} - \ln (\frac{100}{81})$