Integral von $$$- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2$$$

Bestimme $$$\int \left(- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2\right)\, dx$$$.

Gliedweise integrieren:

$${\color{red}{\int{\left(- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2\right)d x}}} = {\color{red}{\left(- \int{2 d x} - \int{x^{2} d x} + \int{e^{\sqrt{2} \sqrt{x}} d x}\right)}}$$

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=2$$$ an:

$$- \int{x^{2} d x} + \int{e^{\sqrt{2} \sqrt{x}} d x} - {\color{red}{\int{2 d x}}} = - \int{x^{2} d x} + \int{e^{\sqrt{2} \sqrt{x}} d x} - {\color{red}{\left(2 x\right)}}$$

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=2$$$ an:

$$- 2 x + \int{e^{\sqrt{2} \sqrt{x}} d x} - {\color{red}{\int{x^{2} d x}}}=- 2 x + \int{e^{\sqrt{2} \sqrt{x}} d x} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- 2 x + \int{e^{\sqrt{2} \sqrt{x}} d x} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Sei $$$u=\sqrt{2} \sqrt{x}$$$.

Dann $$$du=\left(\sqrt{2} \sqrt{x}\right)^{\prime }dx = \frac{\sqrt{2}}{2 \sqrt{x}} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\frac{dx}{\sqrt{x}} = \sqrt{2} du$$$.

Das Integral lässt sich umschreiben als

$$- \frac{x^{3}}{3} - 2 x + {\color{red}{\int{e^{\sqrt{2} \sqrt{x}} d x}}} = - \frac{x^{3}}{3} - 2 x + {\color{red}{\int{u e^{u} d u}}}$$

Für das Integral $$$\int{u e^{u} d u}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{d} \operatorname{dv} = \operatorname{d}\operatorname{v} - \int \operatorname{v} \operatorname{dd}$$$.

Seien $$$\operatorname{d}=u$$$ und $$$\operatorname{dv}=e^{u} du$$$.

Dann gilt $$$\operatorname{dd}=\left(u\right)^{\prime }du=1 du$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (Rechenschritte siehe »).

Also,

$$- \frac{x^{3}}{3} - 2 x + {\color{red}{\int{u e^{u} d u}}}=- \frac{x^{3}}{3} - 2 x + {\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}=- \frac{x^{3}}{3} - 2 x + {\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}$$

Das Integral der Exponentialfunktion lautet $$$\int{e^{u} d u} = e^{u}$$$:

$$u e^{u} - \frac{x^{3}}{3} - 2 x - {\color{red}{\int{e^{u} d u}}} = u e^{u} - \frac{x^{3}}{3} - 2 x - {\color{red}{e^{u}}}$$

Zur Erinnerung: $$$u=\sqrt{2} \sqrt{x}$$$:

$$- \frac{x^{3}}{3} - 2 x - e^{{\color{red}{u}}} + {\color{red}{u}} e^{{\color{red}{u}}} = - \frac{x^{3}}{3} - 2 x - e^{{\color{red}{\sqrt{2} \sqrt{x}}}} + {\color{red}{\sqrt{2} \sqrt{x}}} e^{{\color{red}{\sqrt{2} \sqrt{x}}}}$$

Daher,

$$\int{\left(- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2\right)d x} = \sqrt{2} \sqrt{x} e^{\sqrt{2} \sqrt{x}} - \frac{x^{3}}{3} - 2 x - e^{\sqrt{2} \sqrt{x}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2\right)d x} = \sqrt{2} \sqrt{x} e^{\sqrt{2} \sqrt{x}} - \frac{x^{3}}{3} - 2 x - e^{\sqrt{2} \sqrt{x}}+C$$

$$$\int \left(- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2\right)\, dx = \left(\sqrt{2} \sqrt{x} e^{\sqrt{2} \sqrt{x}} - \frac{x^{3}}{3} - 2 x - e^{\sqrt{2} \sqrt{x}}\right) + C$$$A