Příklad 4.34
[!example] Vypočtěte $\(\Large \int_{0}^{\ln(2)} \frac{5e^x}{15+8e^x+e^{2x}}\;dx\)$
[!tip]+ Substituce
\[\large\begin{aligned} u &= e^x \\ du &= e^x\;dx \\ \end{aligned}\]
\[\large
\begin{align}
&= 5\int_{1}^{2} \frac{1}{15+8u+u^{2}}\;du \\
&= 5\int_{1}^{2} \frac{1}{(u+5)(u+3)}\;du \\
\end{align}
\]
[!tip]+ Rozklad na [[Rozklad na parciální zlomky|parciální zlomky]]
\[\large\begin{aligned} \frac{A}{(u+5)} + \frac{B}{(u+3)} = \begin{cases} A = -\frac{1}{2} \\ B = \frac{1}{2} \end{cases} \\ \end{aligned}\]
\[\large
\begin{align}
&= 5\int_{1}^{2} -\frac{1}{2}\frac{1}{u+5} + \frac{1}{2}\frac{1}{u+3}\;du \\
&= -\frac{5}{2}\int_{1}^{2}\frac{1}{u+5} + \frac{5}{2}\int_{1}^{2}\frac{1}{u+3}\;du \\
&= -\frac{5}{2}\left[\ln(u+5)\right]_{1}^{2} + \frac{5}{2} \left[\ln(u+3)\right]_{1}^{2} \\
&= -\frac{5}{2}\left[\ln(7) - \ln(6)\right] + \frac{5}{2} \left[\ln(5) - \ln(4)\right] \\
&= -\frac{5}{2}\left[\ln\left(\frac{7}{6}\right)\right] + \frac{5}{2} \left[\ln\left(\frac{5}{4}\right)\right] \\
&= \frac{5}{2}\left(\ln\left(\frac{5}{4}\right) - \ln\left(\frac{7}{6}\right)\right) \\
&= \frac{5}{2}\left(\ln\left(\frac{5}{\cancel{4}^2}\cdot\frac{\cancel{6}^3}{7}\right)\right) \\
&= \frac{5}{2}\left(\ln\left(\frac{5}{2}\cdot\frac{3}{7}\right)\right) \\
&= \boxed{\frac{5}{2}\left(\ln\left(\frac{15}{14}\right)\right)} \\
\end{align}
\]