Integral von $$$\frac{u}{u^{2} + 4}$$$

Bestimme $$$\int \frac{u}{u^{2} + 4}\, du$$$.

Sei $$$v=u^{2} + 4$$$.

Dann $$$dv=\left(u^{2} + 4\right)^{\prime }du = 2 u du$$$ (die Schritte sind » zu sehen), und es gilt $$$u du = \frac{dv}{2}$$$.

Somit,

$${\color{red}{\int{\frac{u}{u^{2} + 4} d u}}} = {\color{red}{\int{\frac{1}{2 v} d v}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(v \right)} = \frac{1}{v}$$$ an:

$${\color{red}{\int{\frac{1}{2 v} d v}}} = {\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{2}\right)}}$$

Das Integral von $$$\frac{1}{v}$$$ ist $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$\frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

Zur Erinnerung: $$$v=u^{2} + 4$$$:

$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 4\right)}}}\right| \right)}}{2}$$

Daher,

$$\int{\frac{u}{u^{2} + 4} d u} = \frac{\ln{\left(u^{2} + 4 \right)}}{2}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{u}{u^{2} + 4} d u} = \frac{\ln{\left(u^{2} + 4 \right)}}{2}+C$$

$$$\int \frac{u}{u^{2} + 4}\, du = \frac{\ln\left(u^{2} + 4\right)}{2} + C$$$A