Příklad 4.66

[!example] Vypočtěte $$\Large \int_{-\infty}^{0} \frac{e^{2x}}{1+4e^{4x}}\;dx$$

[!tip]+ Substituce

\[\large\begin{aligned} u &= e^x \\ du &= e^x\;dx \\ \end{aligned}\]

\[\large \begin{align} &= \int_{-\infty}^{0} \frac{e^{x} \cdot e^{x}}{1+4e^{4x}}\;dx \\ &= \int_{0}^{1} \frac{u}{1+4u^4}\;du \\ \end{align} \]

[!tip]+ Substituce

\[\large\begin{aligned} v &= u^2 \\ dv &= 2u\;du \\ \end{aligned}\]

\[\large \begin{align} &= \frac{1}{2}\int_{0}^{1} \frac{2u}{1+4u^4}\;du \\ &= \frac{1}{2}\int_{0}^{1} \frac{1}{1+4v^2}\;dv \\ &= \frac{1}{2}\int_{0}^{1} \frac{1}{1+(2v)^2}\;dv \\ &= \frac{1}{2}\int_{0}^{1} \frac{1}{1+(2v)^2}\;dv \\ \end{align} \]

[!tip]+ Substituce

\[\large\begin{aligned} t &= 2v \\ dt &= 2v\;du \\ \end{aligned}\]

\[\large \begin{align} &= \frac{1}{4}\int_{0}^{1} \frac{2}{1+(2v)^2}\;dv \\ &= \frac{1}{4}\int_{0}^{2} \frac{1}{1+t^2}\;dv \\ &= \frac{1}{4}\left[\arctan(t)\right]_{0}^{2} \\ &= \boxed{\frac{1}{4}\arctan(2)} \\ \end{align} \]