Integral of $$$\frac{\sec{\left(x \right)}}{\tan{\left(x \right)} + \sec{\left(x \right)}}$$$

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

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

Then $$$du=\left(\tan{\left(x \right)} + \sec{\left(x \right)}\right)^{\prime }dx = \left(\tan{\left(x \right)} \sec{\left(x \right)} + \sec^{2}{\left(x \right)}\right) dx$$$ (steps can be seen »), and we have that $$$\left(\tan{\left(x \right)} \sec{\left(x \right)} + \sec^{2}{\left(x \right)}\right) dx = du$$$.

The integral can be rewritten as

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

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

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

$$- {\color{red}{u}}^{-1} = - {\color{red}{\left(\tan{\left(x \right)} + \sec{\left(x \right)}\right)}}^{-1}$$

Therefore,

$$\int{\frac{\sec{\left(x \right)}}{\tan{\left(x \right)} + \sec{\left(x \right)}} d x} = - \frac{1}{\tan{\left(x \right)} + \sec{\left(x \right)}}$$

Add the constant of integration:

$$\int{\frac{\sec{\left(x \right)}}{\tan{\left(x \right)} + \sec{\left(x \right)}} d x} = - \frac{1}{\tan{\left(x \right)} + \sec{\left(x \right)}}+C$$

$$$\int \frac{\sec{\left(x \right)}}{\tan{\left(x \right)} + \sec{\left(x \right)}}\, dx = - \frac{1}{\tan{\left(x \right)} + \sec{\left(x \right)}} + C$$$A