$$$\frac{x - 1}{x^{2} + x + 1}$$$'nin integrali

Bulun: $$$\int \frac{x - 1}{x^{2} + x + 1}\, dx$$$.

Doğrusal terimi $$$x - 1=x\color{red}{+\frac{1}{2}- \frac{1}{2}}-1=x+\frac{1}{2}- \frac{3}{2}$$$ olarak yeniden yazın ve ifadeyi ayırın:

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

Her terimin integralini alın:

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

$$$u=x^{2} + x + 1$$$ olsun.

Böylece $$$du=\left(x^{2} + x + 1\right)^{\prime }dx = \left(2 x + 1\right) dx$$$ (adımlar » görülebilir) ve $$$\left(2 x + 1\right) dx = du$$$ elde ederiz.

İntegral şu şekilde yeniden yazılabilir:

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

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(u \right)} = \frac{1}{u}$$$ ile uygula:

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

$$$\frac{1}{u}$$$'nin integrali $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Hatırlayın ki $$$u=x^{2} + x + 1$$$:

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

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=- \frac{3}{2}$$$ ve $$$f{\left(x \right)} = \frac{1}{x^{2} + x + 1}$$$ ile uygula:

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

Kareye tamamlayın (adımlar » görülebilir): $$$x^{2} + x + 1 = \left(x + \frac{1}{2}\right)^{2} + \frac{3}{4}$$$:

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

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

Böylece $$$du=\left(x + \frac{1}{2}\right)^{\prime }dx = 1 dx$$$ (adımlar » görülebilir) ve $$$dx = du$$$ elde ederiz.

İntegral şu hale gelir

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

$$$v=\frac{2 \sqrt{3} u}{3}$$$ olsun.

Böylece $$$dv=\left(\frac{2 \sqrt{3} u}{3}\right)^{\prime }du = \frac{2 \sqrt{3}}{3} du$$$ (adımlar » görülebilir) ve $$$du = \frac{\sqrt{3} dv}{2}$$$ elde ederiz.

Dolayısıyla,

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

Sabit katsayı kuralı $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$'i $$$c=\frac{2 \sqrt{3}}{3}$$$ ve $$$f{\left(v \right)} = \frac{1}{v^{2} + 1}$$$ ile uygula:

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

$$$\frac{1}{v^{2} + 1}$$$'nin integrali $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

$$\frac{\ln{\left(\left|{x^{2} + x + 1}\right| \right)}}{2} - \sqrt{3} {\color{red}{\int{\frac{1}{v^{2} + 1} d v}}} = \frac{\ln{\left(\left|{x^{2} + x + 1}\right| \right)}}{2} - \sqrt{3} {\color{red}{\operatorname{atan}{\left(v \right)}}}$$

Hatırlayın ki $$$v=\frac{2 \sqrt{3} u}{3}$$$:

$$\frac{\ln{\left(\left|{x^{2} + x + 1}\right| \right)}}{2} - \sqrt{3} \operatorname{atan}{\left({\color{red}{v}} \right)} = \frac{\ln{\left(\left|{x^{2} + x + 1}\right| \right)}}{2} - \sqrt{3} \operatorname{atan}{\left({\color{red}{\left(\frac{2 \sqrt{3} u}{3}\right)}} \right)}$$

Hatırlayın ki $$$u=x + \frac{1}{2}$$$:

$$\frac{\ln{\left(\left|{x^{2} + x + 1}\right| \right)}}{2} - \sqrt{3} \operatorname{atan}{\left(\frac{2 \sqrt{3} {\color{red}{u}}}{3} \right)} = \frac{\ln{\left(\left|{x^{2} + x + 1}\right| \right)}}{2} - \sqrt{3} \operatorname{atan}{\left(\frac{2 \sqrt{3} {\color{red}{\left(x + \frac{1}{2}\right)}}}{3} \right)}$$

Dolayısıyla,

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

Sadeleştirin:

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

İntegrasyon sabitini ekleyin:

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

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