MHB How to solve the integral of 1+tanx using partial fraction decomposition?

[Lorena_Santoro]

\( \int\frac{dx}{1+tanx} \)

 

[I like Serena]

Substitute $u=\tan x$, which means $x=\arctan u$ and $dx=\frac{1}{1+u^2}\,du$.
Follow up with partial fraction decomposition.
$$\begin{align}\int\frac{dx}{1+\tan x}&=\int \frac{1}{1+u}\cdot\frac{1}{1+u^2}\,du
=\int\Big(\frac{\frac 12}{1+u}+\frac{-\frac 12 u +\frac 12}{1+u^2}\Big)\,du
=\frac 12\ln(1+u)-\frac 14\ln(1+u^2)+\frac 12\arctan u + C \\
&=\frac 12\ln(1+\tan x)-\frac 14\ln(1+\tan^2 x) +\frac 12 x + C\end{align}$$

 

[Lorena_Santoro]

Substitute $u=\tan x$, which means $x=\arctan u$ and $dx=\frac{1}{1+u^2}\,du$.
Follow up with partial fraction decomposition.
$$\begin{align}\int\frac{dx}{1+\tan x}&=\int \frac{1}{1+u}\cdot\frac{1}{1+u^2}\,du
=\int\Big(\frac{\frac 12}{1+u}+\frac{-\frac 12 u +\frac 12}{1+u^2}\Big)\,du
=\frac 12\ln(1+u)-\frac 14\ln(1+u^2)+\frac 12\arctan u + C \\
&=\frac 12\ln(1+\tan x)-\frac 14\ln(1+\tan^2 x) +\frac 12 x + C\end{align}$$

Thank you very much!