Integral of $$$\frac{1}{1 - \tan{\left(x \right)}}$$$

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

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

Then $$$x=\operatorname{atan}{\left(u \right)}$$$ and $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (steps can be seen »).

So,

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

Perform partial fraction decomposition (steps can be seen »):

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

Integrate term by term:

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \frac{1}{u - 1}$$$:

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

Let $$$v=u - 1$$$.

Then $$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (steps can be seen »), and we have that $$$du = dv$$$.

Therefore,

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

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Recall that $$$v=u - 1$$$:

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \frac{u + 1}{u^{2} + 1}$$$:

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

Split the fraction:

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

Integrate term by term:

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

Let $$$v=u^{2} + 1$$$.

Then $$$dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$$$ (steps can be seen »), and we have that $$$u du = \frac{dv}{2}$$$.

So,

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

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(v \right)} = \frac{1}{v}$$$:

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

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Recall that $$$v=u^{2} + 1$$$:

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

The integral of $$$\frac{1}{u^{2} + 1}$$$ is $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

Recall that $$$u=\tan{\left(x \right)}$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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