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

Man berechne:

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

## Lösungsvorschlag

### Substitution:

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

### Partialbruchzerlegung:

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

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

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

$A = B = 1 \qquad C=-1$

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

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

### Integrieren:

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

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

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

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

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

### Rücksubstitution:

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