Příklad 4.61
[!example] Vypočtěte $\(\Large \int_{1}^{e} 4x\cdot\ln(2x)\;dx\)$
\[\large
\begin{align}
&= \int_{1}^{e} 2x\cdot\ln(2x)\;2dx \\
\end{align}
\]
[!tip]+ Substituce
\[\large\begin{aligned} u &= 2x \\ du &= 2dx \\ \end{aligned}\]
\[\large
\begin{align}
&= \int_{2}^{2e} u\cdot\ln(u)\;du \\
\end{align}
\]
[!tip]+ Aplikace [[Per Partes#Metoda DI|DI metody]] Sestavíme si DI tabulku:
D I + \(\ln(u)\) \(u\) - \(\frac{1}{u}\) \(\frac{u^2}{2}\) Zapíšeme výsledek: \(\frac{1}{2}\ln(u)u^2-\int_2^{2e}\frac{u}{2}\;du\)
\[\large
\begin{align}
&= \left[\frac{1}{2}\ln(u)u^2\right]_{2}^{2e}-\frac{1}{2}\int_2^{2e}u\;du \\
&= \left[\frac{1}{2}\ln(u)u^2\right]_{2}^{2e}-\frac{1}{2}\left[\frac{u^2}{2}\right]_{2}^{2e} \\
&= \frac{1}{2}\left(4e^2\ln(2e) - 4\ln(2)\right)-\frac{1}{2}\left[\frac{4e^2 - 4}{2}\right] \\
&= \frac{1}{2}\left(4e^2\ln(2e) - 4\ln(2)\right)-\frac{1}{2}\left[{2e^2 - 2}\right] \\
&= \left(2e^2(\ln(2) + \ln(e)) - 2\ln(2)\right)-e^2 + 1 \\
&= 2e^2\ln(2) + 2e^2 - 2\ln(2)-e^2 + 1 \\
&= 2e^2\ln(2) - 2\ln(2) + 1 + e^2 \\
&= \boxed{\ln(2)\cdot(2e^2 - 2) + e^2 + 1}
\end{align}
\]