(a) Find: ∫ (2 x-1)/((x-1)(x+2)(x-3)) d x

\[\begin{array}{r} \int \frac{2 x-1}{(x-1)(x+2)(x-3)} \mathrm{d} x=-\frac{1}{6} \int \frac{1}{x-1} \mathrm{d} x-\frac{1}{3} \int \frac{1}{x+2} \mathrm{d} x+\frac{1}{2} \int \frac{1}{x-3} \mathrm{d} x \\ \text { (Using Partial Fraction) } \\ =-\frac{1}{6} \log |x-1|-\frac{1}{3} \log |x+2|+\frac{1}{2} \log |x-3|+C \end{array} \] \[\begin{aligned} & I=\int_{0}^{5}(|x-1|+|x-2|+|x-5|) \mathrm{d} x \\ & \therefore I=\left[-\int_{0}^{1}(x-1) \mathrm{d} x+\int_{1}^{5}(x-1) \mathrm{d} x\right]+\left[-\int_{0}^{2}(x-2) \mathrm{d} x+\int_{2}^{5}(x-2) \mathrm{d} x\right]+\left[-\int_{0}^{5}(x-5) \mathrm{d} x\right] \\ & \left.\left.\left.\left.\left.=-\frac{(x-1)^{2}}{2}\right]_{0}^{1}+\frac{(x-1)^{2}}{2}\right]_{1}^{5}-\frac{(x-2)^{2}}{2}\right]_{0}^{2}+\frac{(x-2)^{2}}{2}\right]_{2}^{5}-\frac{(x-5)^{2}}{2}\right]_{0}^{5} \\ & =\frac{17}{2}+\frac{13}{2}+\frac{25}{2}=\frac{55}{2} \end{aligned} \]