قرار میدهیم $u=2+tanx $ ,در نتیجه $du=(1+tan^2x)dx $ داریم:
$$\begin{align} \int \frac{dx}{2+tanx}&= \int \frac{du}{[1+(u-2)^2]u}\\
&=- \frac{1}{5} \int \frac{u-1}{(u-1)^2+4}+ \frac{1}{5} \int \frac{1}{(u-1)^2+4} \\
&+ \frac{1}{5} \int \frac{1}{u}du\\
&= - \frac{1}{10}ln | (u-1)^2+4 | + \frac{1}{20}tan^-1( \frac{u-1}{2} )^2\\
&+ \frac{1}{5}ln | u | \end{align}$$
