Integral de $$$\sec^{2}{\left(x + 1 \right)}$$$

Halla $$$\int \sec^{2}{\left(x + 1 \right)}\, dx$$$.

Sea $$$u=x + 1$$$.

Entonces $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.

Por lo tanto,

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

La integral de $$$\sec^{2}{\left(u \right)}$$$ es $$$\int{\sec^{2}{\left(u \right)} d u} = \tan{\left(u \right)}$$$:

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

Recordemos que $$$u=x + 1$$$:

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

Por lo tanto,

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

Añade la constante de integración:

$$\int{\sec^{2}{\left(x + 1 \right)} d x} = \tan{\left(x + 1 \right)}+C$$

$$$\int \sec^{2}{\left(x + 1 \right)}\, dx = \tan{\left(x + 1 \right)} + C$$$A