Integral dari $$$\frac{1}{\tan^{2}{\left(x \right)} - 1}$$$

Temukan $$$\int \frac{1}{\tan^{2}{\left(x \right)} - 1}\, dx$$$.

Misalkan $$$u=\tan{\left(x \right)}$$$.

Kemudian $$$x=\operatorname{atan}{\left(u \right)}$$$ dan $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (langkah-langkahnya dapat dilihat »).

Dengan demikian,

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

Lakukan dekomposisi pecahan parsial (langkah-langkah dapat dilihat di »):

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

Integralkan suku demi suku:

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$:

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

Integral dari $$$\frac{1}{u^{2} + 1}$$$ adalah $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$\int{\frac{1}{4 \left(u - 1\right)} d u} - \int{\frac{1}{4 \left(u + 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{2} = \int{\frac{1}{4 \left(u - 1\right)} d u} - \int{\frac{1}{4 \left(u + 1\right)} d u} - \frac{{\color{red}{\operatorname{atan}{\left(u \right)}}}}{2}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{4}$$$ dan $$$f{\left(u \right)} = \frac{1}{u + 1}$$$:

$$- \frac{\operatorname{atan}{\left(u \right)}}{2} + \int{\frac{1}{4 \left(u - 1\right)} d u} - {\color{red}{\int{\frac{1}{4 \left(u + 1\right)} d u}}} = - \frac{\operatorname{atan}{\left(u \right)}}{2} + \int{\frac{1}{4 \left(u - 1\right)} d u} - {\color{red}{\left(\frac{\int{\frac{1}{u + 1} d u}}{4}\right)}}$$

Misalkan $$$v=u + 1$$$.

Kemudian $$$dv=\left(u + 1\right)^{\prime }du = 1 du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$du = dv$$$.

Integral tersebut dapat ditulis ulang sebagai

$$- \frac{\operatorname{atan}{\left(u \right)}}{2} + \int{\frac{1}{4 \left(u - 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{u + 1} d u}}}}{4} = - \frac{\operatorname{atan}{\left(u \right)}}{2} + \int{\frac{1}{4 \left(u - 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4}$$

Integral dari $$$\frac{1}{v}$$$ adalah $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$- \frac{\operatorname{atan}{\left(u \right)}}{2} + \int{\frac{1}{4 \left(u - 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = - \frac{\operatorname{atan}{\left(u \right)}}{2} + \int{\frac{1}{4 \left(u - 1\right)} d u} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$

Ingat bahwa $$$v=u + 1$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} - \frac{\operatorname{atan}{\left(u \right)}}{2} + \int{\frac{1}{4 \left(u - 1\right)} d u} = - \frac{\ln{\left(\left|{{\color{red}{\left(u + 1\right)}}}\right| \right)}}{4} - \frac{\operatorname{atan}{\left(u \right)}}{2} + \int{\frac{1}{4 \left(u - 1\right)} d u}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{4}$$$ dan $$$f{\left(u \right)} = \frac{1}{u - 1}$$$:

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

Misalkan $$$v=u - 1$$$.

Kemudian $$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$du = dv$$$.

Integralnya menjadi

$$- \frac{\ln{\left(\left|{u + 1}\right| \right)}}{4} - \frac{\operatorname{atan}{\left(u \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{u - 1} d u}}}}{4} = - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{4} - \frac{\operatorname{atan}{\left(u \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4}$$

Integral dari $$$\frac{1}{v}$$$ adalah $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$- \frac{\ln{\left(\left|{u + 1}\right| \right)}}{4} - \frac{\operatorname{atan}{\left(u \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{4} - \frac{\operatorname{atan}{\left(u \right)}}{2} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$

Ingat bahwa $$$v=u - 1$$$:

$$- \frac{\ln{\left(\left|{u + 1}\right| \right)}}{4} + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} - \frac{\operatorname{atan}{\left(u \right)}}{2} = - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{4} + \frac{\ln{\left(\left|{{\color{red}{\left(u - 1\right)}}}\right| \right)}}{4} - \frac{\operatorname{atan}{\left(u \right)}}{2}$$

Ingat bahwa $$$u=\tan{\left(x \right)}$$$:

$$\frac{\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)}}{4} - \frac{\ln{\left(\left|{1 + {\color{red}{u}}}\right| \right)}}{4} - \frac{\operatorname{atan}{\left({\color{red}{u}} \right)}}{2} = \frac{\ln{\left(\left|{-1 + {\color{red}{\tan{\left(x \right)}}}}\right| \right)}}{4} - \frac{\ln{\left(\left|{1 + {\color{red}{\tan{\left(x \right)}}}}\right| \right)}}{4} - \frac{\operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)}}{2}$$

Oleh karena itu,

$$\int{\frac{1}{\tan^{2}{\left(x \right)} - 1} d x} = \frac{\ln{\left(\left|{\tan{\left(x \right)} - 1}\right| \right)}}{4} - \frac{\ln{\left(\left|{\tan{\left(x \right)} + 1}\right| \right)}}{4} - \frac{\operatorname{atan}{\left(\tan{\left(x \right)} \right)}}{2}$$

Sederhanakan:

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

Tambahkan konstanta integrasi:

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

$$$\int \frac{1}{\tan^{2}{\left(x \right)} - 1}\, dx = \left(- \frac{x}{2} + \frac{\ln\left(\left|{\tan{\left(x \right)} - 1}\right|\right)}{4} - \frac{\ln\left(\left|{\tan{\left(x \right)} + 1}\right|\right)}{4}\right) + C$$$A