jeszcze jedna całeczka

[speedy1]

nie wychodza mi te wzorki na calki mam problem z calka e^x przez [e^2x + 2]

[Nakahed90]

\(\displaystyle{ t=e^{x}}\)
\(\displaystyle{ dt=e^{x}dx}\)
\(\displaystyle{ =\int \frac{dt}{t^{2}+(\sqrt{2})^{2}}=\frac{\sqrt{2}}{2}arctg(\frac{\sqrt{2}e^{x}}{2})+C}\)

[sea_of_tears]

\(\displaystyle{ \int_{}^{} \frac{ e^{x} }{ e^{2x}+2 } dx=\newline
\begin{cases}
t=e^x \\
dt=e^x dx
\end{cases}
\newline
=\int \frac{dt}{t^2+2}= \newline
\begin{cases}
t=\sqrt2u \\
dt=\sqrt2 du
\end{cases}
\newline
=\sqrt{2} \int\frac{du}{2u^2+2}=
\frac{\sqrt2}{2}\int \frac{du}{u^2+1}=\newline
=\frac{\sqrt2}{2}arctgu+c=
\frac{\sqrt2}{2}arctg(\frac{t}{\sqrt2})+c=
\frac{\sqrt2}{2}arctg(\frac{e^x}{\sqrt2})+c}\)

[gufox]

\(\displaystyle{ \int \frac{e ^{x}dx }{e ^{2x}+2 } = \begin{cases} e ^{x}=t \\ e ^{x}dx=dt \end{cases}=\int \frac{dt}{t ^{2}+2 }= \frac{1}{ \sqrt{2} }arctg \frac{t}{ \sqrt{2} }+C=...}\)