Integral von $$$\frac{1}{\sqrt{x^{2} + x + 1}}$$$

Bestimme $$$\int \frac{1}{\sqrt{x^{2} + x + 1}}\, dx$$$.

Quadrat ergänzen (Schritte siehe »): $$$x^{2} + x + 1 = \left(x + \frac{1}{2}\right)^{2} + \frac{3}{4}$$$:

$${\color{red}{\int{\frac{1}{\sqrt{x^{2} + x + 1}} d x}}} = {\color{red}{\int{\frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^{2} + \frac{3}{4}}} d x}}}$$

Sei $$$u=x + \frac{1}{2}$$$.

Dann $$$du=\left(x + \frac{1}{2}\right)^{\prime }dx = 1 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = du$$$.

Das Integral lässt sich umschreiben als

$${\color{red}{\int{\frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^{2} + \frac{3}{4}}} d x}}} = {\color{red}{\int{\frac{1}{\sqrt{u^{2} + \frac{3}{4}}} d u}}}$$

Sei $$$u=\frac{\sqrt{3} \sinh{\left(v \right)}}{2}$$$.

Dann $$$du=\left(\frac{\sqrt{3} \sinh{\left(v \right)}}{2}\right)^{\prime }dv = \frac{\sqrt{3} \cosh{\left(v \right)}}{2} dv$$$ (die Schritte sind » zu sehen).

Somit folgt, dass $$$v=\operatorname{asinh}{\left(\frac{2 \sqrt{3} u}{3} \right)}$$$.

Daher,

$$$\frac{1}{\sqrt{ u ^{2} + \frac{3}{4}}} = \frac{1}{\sqrt{\frac{3 \sinh^{2}{\left( v \right)}}{4} + \frac{3}{4}}}$$$

Verwenden Sie die Identität $$$\sinh^{2}{\left( v \right)} + 1 = \cosh^{2}{\left( v \right)}$$$:

$$$\frac{1}{\sqrt{\frac{3 \sinh^{2}{\left( v \right)}}{4} + \frac{3}{4}}}=\frac{2 \sqrt{3}}{3 \sqrt{\sinh^{2}{\left( v \right)} + 1}}=\frac{2 \sqrt{3}}{3 \sqrt{\cosh^{2}{\left( v \right)}}}$$$

$$$\frac{2 \sqrt{3}}{3 \sqrt{\cosh^{2}{\left( v \right)}}} = \frac{2 \sqrt{3}}{3 \cosh{\left( v \right)}}$$$

Das Integral wird zu

$${\color{red}{\int{\frac{1}{\sqrt{u^{2} + \frac{3}{4}}} d u}}} = {\color{red}{\int{1 d v}}}$$

Wenden Sie die Konstantenregel $$$\int c\, dv = c v$$$ mit $$$c=1$$$ an:

$${\color{red}{\int{1 d v}}} = {\color{red}{v}}$$

Zur Erinnerung: $$$v=\operatorname{asinh}{\left(\frac{2 \sqrt{3} u}{3} \right)}$$$:

$${\color{red}{v}} = {\color{red}{\operatorname{asinh}{\left(\frac{2 \sqrt{3} u}{3} \right)}}}$$

Zur Erinnerung: $$$u=x + \frac{1}{2}$$$:

$$\operatorname{asinh}{\left(\frac{2 \sqrt{3} {\color{red}{u}}}{3} \right)} = \operatorname{asinh}{\left(\frac{2 \sqrt{3} {\color{red}{\left(x + \frac{1}{2}\right)}}}{3} \right)}$$

Daher,

$$\int{\frac{1}{\sqrt{x^{2} + x + 1}} d x} = \operatorname{asinh}{\left(\frac{2 \sqrt{3} \left(x + \frac{1}{2}\right)}{3} \right)}$$

Vereinfachen:

$$\int{\frac{1}{\sqrt{x^{2} + x + 1}} d x} = \operatorname{asinh}{\left(\frac{\sqrt{3} \left(2 x + 1\right)}{3} \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sqrt{x^{2} + x + 1}} d x} = \operatorname{asinh}{\left(\frac{\sqrt{3} \left(2 x + 1\right)}{3} \right)}+C$$

$$$\int \frac{1}{\sqrt{x^{2} + x + 1}}\, dx = \operatorname{asinh}{\left(\frac{\sqrt{3} \left(2 x + 1\right)}{3} \right)} + C$$$A