TU Wien:Analysis UE (diverse)/Übungen SS19/Beispiel 232

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Man berechne:

\int{\frac{(e^x-1)}{(e^{2x}+1)}}dx

Lösungsvorschlag[Bearbeiten]

Substitution:[Bearbeiten]


e^x = u ;
dx*e^x = du ;
 dx = \frac{du}{u}

Partialbruchzerlegung:[Bearbeiten]


\frac{u-1}{u(u^2+1)}
= \frac{Au+B}{u^2+1} + \frac{C}{u} \quad | \cdot u (u^2 +1)


u-1 = (A u+B)u + C(u^2 +1)
= A u^2 + B u + C(u^2 +1)

u-1 = 1 u^2 + 1 u + (-1) (u^2 +1)

A = B = 1 \qquad C=-1


\frac{Au+B}{u^2+1} + \frac{C}{u}
= \frac{u+1}{u^2+1} - \frac{1}{u}
= \frac{u}{u^2+1} + \frac{1}{u^2+1} - \frac{1}{u}


\int ( \frac{u}{u^2+1} + \frac{1}{u^2+1} - \frac{1}{u} ) du

Integrieren:[Bearbeiten]

\int \frac{u \, du}{u^2+1} + \int \frac{du}{u^2+1} - \int \frac{du}{u}


\int \frac{u \, du}{u^2+1} \quad : \qquad
v:=u^2+1 ; \quad
dv=2u\,du \quad : \qquad
\frac{1}{2}\int \frac{dv}{v} = \frac{1}{2}\ln v \quad : \qquad
\frac{1}{2}\int \frac{u \, du}{u^2+1} = \frac{1}{2}\ln (u^2+1)

\int \frac{du}{u^2+1} = \arctan(u)

\int \frac{du}{u} = \ln u

\int \frac{u \, du}{u^2+1} + \int \frac{du}{u^2+1} - \int \frac{du}{u} = \frac{1}{2}\ln (u^2+1) + \arctan u - \ln u

Rücksubstitution:[Bearbeiten]

= \frac{1}{2}\ln (e^{2x} + 1) + \arctan e^x - x