Příklad 4.55

\[\large \begin{align} &= \left[\ln(4x)\cdot\left(\frac{x^2}{2}-2x\right)\right]_{1}^{2} - \int_{1}^{2}\frac{1}{\cancel{x}^1}\cdot\left(\frac{\cancel{x^2}^x}{2}-2\cancel{x}^1\right)\;dx \\ &= \left[\ln(4x)\cdot\left(\frac{x^2}{2}-2x\right)\right]_{1}^{2} - \frac{1}{2}\int_{1}^{2}x\;dx + 2\int_{1}^{2}1\;dx \\ &= \left[\ln(4x)\cdot\left(\frac{x^2}{2}-2x\right) - \frac{x^2}{4}\right]_{1}^{2} + 2\int_{1}^{2}1\;dx \\ &= \left[\ln(4x)\cdot\left(\frac{x^2}{2}-2x\right) - \frac{1}{2}\cdot\frac{x^2}{2}\right]_{1}^{2} + 2\int_{1}^{2}1\;dx \\ &= \left[\ln(4x)\cdot\left(\frac{x^2}{2}-2x\right) - \frac{x^2}{4} + 2x\right]_{1}^{2}\\ &= \left[\ln(8)\cdot\left(\frac{4}{2}-4\right) - \frac{4}{4} + 4\right] - \left[\ln(4)\cdot\left(\frac{1}{2}-2\right) - \frac{1}{4} + 2\right]\\ &= \left[-2\ln(8) + 3\right] - \left[-\frac{3}{2}\ln(4)-\frac{1}{4} + \frac{8}{4}\right] \\ &= -2\ln\left(2^3\right) + 3 + \frac{3}{2}\ln\left(2^2\right) - \frac{7}{4} \\ &= -6\ln(2) + 3 + \frac{6}{2}\ln(2) - \frac{7}{4} \\ &= -6\ln(2) + \frac{12}{4} + 3\ln(2) - \frac{7}{4} \\ &= \boxed{-3\ln(2) + \frac{5}{4}} \end{align} \]