Integral de $$$\frac{1}{x^{2} + 2 x + 2}$$$

Halla $$$\int \frac{1}{x^{2} + 2 x + 2}\, dx$$$.

Completa el cuadrado (se pueden ver los pasos »): $$$x^{2} + 2 x + 2 = \left(x + 1\right)^{2} + 1$$$:

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

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

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

La integral puede reescribirse como

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

La integral de $$$\frac{1}{u^{2} + 1}$$$ es $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

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

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

Por lo tanto,

$$\int{\frac{1}{x^{2} + 2 x + 2} d x} = \operatorname{atan}{\left(x + 1 \right)}$$

Añade la constante de integración:

$$\int{\frac{1}{x^{2} + 2 x + 2} d x} = \operatorname{atan}{\left(x + 1 \right)}+C$$

$$$\int \frac{1}{x^{2} + 2 x + 2}\, dx = \operatorname{atan}{\left(x + 1 \right)} + C$$$A