Příklad 4.60

[!example] Vypočtěte $\(\Large \int_{\ln{3}}^{\ln{5}} \frac{e^{3x}+e^{2x}}{e^{2x}-4e^x+4}\;dx\)$

\[\large \begin{align} &= \int_{\ln{3}}^{\ln{5}} \frac{e^x\cdot(e^{2x}+e^{x})}{e^{2x}-4e^x+4}\;dx \\ \end{align} \]

[!tip]+ Substituce

\[\large\begin{aligned} u &= e^x \\ du &= e^xdx \\ \end{aligned}\]
\[\large \begin{align} &= \int_{3}^{5} \frac{u^2+u}{u^2-4u+4}\;du \\ \end{align} \]

[!tip]+ Dělení polynomů

\[\large\begin{aligned} (u^2+u) &: (u^2-4u+4) = 1 + \frac{5u-4}{u^2-4u+4}\\ -(u^2-4u+4) \\ \\ 5u-4 \\ \end{aligned}\]
\[\large \begin{align} &= \int_{3}^{5} 1+ \frac{5u-4}{u^2-4u+4}\;du \\ &= \int_{3}^{5} 1\;du + \int_{3}^{5}\frac{5u-4}{u^2-4u+4}\;du \\ &= \int_{3}^{5} 1\;du + \int_{3}^{5}\frac{5u-4}{u^2-4u+4}\;du \\ &= \left[u\right]_{3}^{5} + 5\int_{3}^{5}\frac{u}{(u-2)^2}\;du - 4\int_{3}^{5}\frac{1}{(u-2)^2}\;du \\ \end{align} \]

[!tip]+ Substituce

\[\large\begin{aligned} v &= u-2 \\ dv &= du \\ \end{aligned}\]
\[\large \begin{align} &= \left[u\right]_{3}^{5} + 5\int_{1}^{3}\frac{v+2}{v^2}\;dv - 4\int_{1}^{3}\frac{1}{v^2}\;du \\ &= \left[u\right]_{3}^{5} + 5\int_{1}^{3}\frac{v}{v^2}\;dv + 10\int_{1}^{3}\frac{1}{v^2}\;dv - 4\int_{1}^{3}\frac{1}{v^2}\;du \\ &= \left[u\right]_{3}^{5} + 5\int_{1}^{3}\frac{1}{v}\;dv + 6\int_{1}^{3}\frac{1}{v^2}\;dv\\ &= \left[u\right]_{3}^{5} + 5\left[\;\ln|u|\;\right]_{1}^{3} + 6\left[\;-\frac{1}{v}\;\right]_{1}^{3}\\ &= 2 + 5[\ln(3)-\ln(1)] + 6\left[\;-\frac{1}{3} + 1\;\right]_{1}^{3}\\ &= 2 + 5\ln(3) + -2 + 6\\ &= \boxed{6 + 5\ln(3)}\\ \end{align} \]