Integral de $$$\frac{1}{y^{2} - 3}$$$

Encontre $$$\int \frac{1}{y^{2} - 3}\, dy$$$.

Efetue a decomposição em frações parciais (os passos podem ser vistos »):

$${\color{red}{\int{\frac{1}{y^{2} - 3} d y}}} = {\color{red}{\int{\left(- \frac{\sqrt{3}}{6 \left(y + \sqrt{3}\right)} + \frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)}\right)d y}}}$$

Integre termo a termo:

$${\color{red}{\int{\left(- \frac{\sqrt{3}}{6 \left(y + \sqrt{3}\right)} + \frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)}\right)d y}}} = {\color{red}{\left(\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \int{\frac{\sqrt{3}}{6 \left(y + \sqrt{3}\right)} d y}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ usando $$$c=\frac{\sqrt{3}}{6}$$$ e $$$f{\left(y \right)} = \frac{1}{y + \sqrt{3}}$$$:

$$\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - {\color{red}{\int{\frac{\sqrt{3}}{6 \left(y + \sqrt{3}\right)} d y}}} = \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - {\color{red}{\left(\frac{\sqrt{3} \int{\frac{1}{y + \sqrt{3}} d y}}{6}\right)}}$$

Seja $$$u=y + \sqrt{3}$$$.

Então $$$du=\left(y + \sqrt{3}\right)^{\prime }dy = 1 dy$$$ (veja os passos »), e obtemos $$$dy = du$$$.

Logo,

$$\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{y + \sqrt{3}} d y}}}}{6} = \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6}$$

A integral de $$$\frac{1}{u}$$$ é $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6} = \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \frac{\sqrt{3} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

Recorde que $$$u=y + \sqrt{3}$$$:

$$- \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{6} + \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} = - \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{\left(y + \sqrt{3}\right)}}}\right| \right)}}{6} + \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ usando $$$c=\frac{\sqrt{3}}{6}$$$ e $$$f{\left(y \right)} = \frac{1}{y - \sqrt{3}}$$$:

$$- \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + {\color{red}{\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y}}} = - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + {\color{red}{\left(\frac{\sqrt{3} \int{\frac{1}{y - \sqrt{3}} d y}}{6}\right)}}$$

Seja $$$u=y - \sqrt{3}$$$.

Então $$$du=\left(y - \sqrt{3}\right)^{\prime }dy = 1 dy$$$ (veja os passos »), e obtemos $$$dy = du$$$.

A integral pode ser reescrita como

$$- \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{y - \sqrt{3}} d y}}}}{6} = - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6}$$

A integral de $$$\frac{1}{u}$$$ é $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6} = - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

Recorde que $$$u=y - \sqrt{3}$$$:

$$- \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{6} = - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{\left(y - \sqrt{3}\right)}}}\right| \right)}}{6}$$

Portanto,

$$\int{\frac{1}{y^{2} - 3} d y} = \frac{\sqrt{3} \ln{\left(\left|{y - \sqrt{3}}\right| \right)}}{6} - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6}$$

Simplifique:

$$\int{\frac{1}{y^{2} - 3} d y} = \frac{\sqrt{3} \left(\ln{\left(\left|{y - \sqrt{3}}\right| \right)} - \ln{\left(\left|{y + \sqrt{3}}\right| \right)}\right)}{6}$$

Adicione a constante de integração:

$$\int{\frac{1}{y^{2} - 3} d y} = \frac{\sqrt{3} \left(\ln{\left(\left|{y - \sqrt{3}}\right| \right)} - \ln{\left(\left|{y + \sqrt{3}}\right| \right)}\right)}{6}+C$$

$$$\int \frac{1}{y^{2} - 3}\, dy = \frac{\sqrt{3} \left(\ln\left(\left|{y - \sqrt{3}}\right|\right) - \ln\left(\left|{y + \sqrt{3}}\right|\right)\right)}{6} + C$$$A