Find $\int_{1}^{2}\dfrac{3x^{3}-6x^{2}-5x+4}{2x}dx$.

Solution The function $f(x) =\dfrac{3x^{3}-6x^{2}-5x+4}{2x}$ is continuous on the closed interval $[1,2]$. Using algebra and the properties of the definite integral, we get $$\begin{eqnarray*} \int_{1}^{2}\dfrac{3x^{3}-6x^{2}-5x+4}{2x}dx &=&\int_{1}^{2}\left[\dfrac{3}{2} x^{2}- 3x- \dfrac{5}{2}+\dfrac{2}{x}\right] dx\\ &=& \int_{1}^{2} \dfrac{3}{2} x^{2}dx-\int_{1}^{2} 3x\, dx- \int_{1}^{2} \dfrac{5}{2} \, dx+\int_{1}^{2}\dfrac{2}{x}~dx \\[4pt] &=& \dfrac{3}{2}\int_{1}^{2}x^{2}~dx-3\int_{1}^{2}x\, dx-\dfrac{5}{2}\int_{1}^{2}~dx+2\int_{1}^{2}\dfrac{1}{x}~dx \\ &=& \dfrac{3}{2} \left[ \dfrac{x^{3}}{3}\right] _{1}^{2} -3 \left[ \dfrac{x^{2}}{2}\right] _{1}^{2} -\dfrac{5}{2} \Big[x\Big]_{1}^{2} +2 \Big[\ln \vert x\vert \Big]_{1}^{2} \\ &=&\dfrac{1}{2} ( 8-1) - 3\left( 2-\dfrac{1}{2}\right) -\dfrac{5}{2}(2-1) + 2( \ln 2- \ln 1 ) \\ &=&-\dfrac{7}{2}+2\ln 2\approx -2.114 \end{eqnarray*}$$