Integral of (1+tg(x)+ctg(x)) dx

1. Integrate term-by-term:

   1. Integrate term-by-term:

      1. Rewrite the integrand:

         $$\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$$

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

         Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

         $$\int \left(- \frac{1}{u}\right)\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            So, the result is: $- \log{\left(u \right)}$

         Now substitute $u$ back in:

         $$- \log{\left(\cos{\left(x \right)} \right)}$$

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dx = x$$

      The result is: $x - \log{\left(\cos{\left(x \right)} \right)}$

   1. Rewrite the integrand:

      $$\cot{\left(x \right)} = \frac{\cos{\left(x \right)}}{\sin{\left(x \right)}}$$

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

      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

      $$\int \frac{1}{u}\, du$$

      1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

      Now substitute $u$ back in:

      $$\log{\left(\sin{\left(x \right)} \right)}$$

   The result is: $x + \log{\left(\sin{\left(x \right)} \right)} - \log{\left(\cos{\left(x \right)} \right)}$

2. Add the constant of integration:

   $$x + \log{\left(\sin{\left(x \right)} \right)} - \log{\left(\cos{\left(x \right)} \right)}+ \mathrm{constant}$$

The answer is:

$$x + \log{\left(\sin{\left(x \right)} \right)} - \log{\left(\cos{\left(x \right)} \right)}+ \mathrm{constant}$$