Integral von $$$22 i a^{2} b^{x - 1} n t x$$$ nach $$$x$$$

Bestimme $$$\int 22 i a^{2} b^{x - 1} n t x\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=22 i a^{2} n t$$$ und $$$f{\left(x \right)} = b^{x - 1} x$$$ an:

$${\color{red}{\int{22 i a^{2} b^{x - 1} n t x d x}}} = {\color{red}{\left(22 i a^{2} n t \int{b^{x - 1} x d x}\right)}}$$

Für das Integral $$$\int{b^{x - 1} x d x}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Seien $$$\operatorname{u}=x$$$ und $$$\operatorname{dv}=b^{x - 1} dx$$$.

Dann gilt $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{b^{x - 1} d x}=\frac{b^{x - 1}}{\ln{\left(b \right)}}$$$ (Rechenschritte siehe »).

Also,

$$22 i a^{2} n t {\color{red}{\int{b^{x - 1} x d x}}}=22 i a^{2} n t {\color{red}{\left(x \cdot \frac{b^{x - 1}}{\ln{\left(b \right)}}-\int{\frac{b^{x - 1}}{\ln{\left(b \right)}} \cdot 1 d x}\right)}}=22 i a^{2} n t {\color{red}{\left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \int{\frac{b^{x - 1}}{\ln{\left(b \right)}} d x}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{\ln{\left(b \right)}}$$$ und $$$f{\left(x \right)} = b^{x - 1}$$$ an:

$$22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - {\color{red}{\int{\frac{b^{x - 1}}{\ln{\left(b \right)}} d x}}}\right) = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - {\color{red}{\frac{\int{b^{x - 1} d x}}{\ln{\left(b \right)}}}}\right)$$

Sei $$$u=x - 1$$$.

Dann $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = du$$$.

Daher,

$$22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{{\color{red}{\int{b^{x - 1} d x}}}}{\ln{\left(b \right)}}\right) = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{{\color{red}{\int{b^{u} d u}}}}{\ln{\left(b \right)}}\right)$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=b$$$:

$$22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{{\color{red}{\int{b^{u} d u}}}}{\ln{\left(b \right)}}\right) = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{{\color{red}{\frac{b^{u}}{\ln{\left(b \right)}}}}}{\ln{\left(b \right)}}\right)$$

Zur Erinnerung: $$$u=x - 1$$$:

$$22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{b^{{\color{red}{u}}}}{\ln{\left(b \right)}^{2}}\right) = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{b^{{\color{red}{\left(x - 1\right)}}}}{\ln{\left(b \right)}^{2}}\right)$$

Daher,

$$\int{22 i a^{2} b^{x - 1} n t x d x} = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{b^{x - 1}}{\ln{\left(b \right)}^{2}}\right)$$

Vereinfachen:

$$\int{22 i a^{2} b^{x - 1} n t x d x} = \frac{22 i a^{2} b^{x - 1} n t \left(x \ln{\left(b \right)} - 1\right)}{\ln{\left(b \right)}^{2}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{22 i a^{2} b^{x - 1} n t x d x} = \frac{22 i a^{2} b^{x - 1} n t \left(x \ln{\left(b \right)} - 1\right)}{\ln{\left(b \right)}^{2}}+C$$

$$$\int 22 i a^{2} b^{x - 1} n t x\, dx = \frac{22 i a^{2} b^{x - 1} n t \left(x \ln\left(b\right) - 1\right)}{\ln^{2}\left(b\right)} + C$$$A