Integral de $$$11 x + \frac{17}{2 x^{2} + 7 x - 4}$$$

Halla $$$\int \left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)\, dx$$$.

Integra término a término:

$${\color{red}{\int{\left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)d x}}} = {\color{red}{\left(\int{11 x d x} + \int{\frac{17}{2 x^{2} + 7 x - 4} d x}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=11$$$ y $$$f{\left(x \right)} = x$$$:

$$\int{\frac{17}{2 x^{2} + 7 x - 4} d x} + {\color{red}{\int{11 x d x}}} = \int{\frac{17}{2 x^{2} + 7 x - 4} d x} + {\color{red}{\left(11 \int{x d x}\right)}}$$

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=1$$$:

$$\int{\frac{17}{2 x^{2} + 7 x - 4} d x} + 11 {\color{red}{\int{x d x}}}=\int{\frac{17}{2 x^{2} + 7 x - 4} d x} + 11 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\int{\frac{17}{2 x^{2} + 7 x - 4} d x} + 11 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=17$$$ y $$$f{\left(x \right)} = \frac{1}{2 x^{2} + 7 x - 4}$$$:

$$\frac{11 x^{2}}{2} + {\color{red}{\int{\frac{17}{2 x^{2} + 7 x - 4} d x}}} = \frac{11 x^{2}}{2} + {\color{red}{\left(17 \int{\frac{1}{2 x^{2} + 7 x - 4} d x}\right)}}$$

Realizar la descomposición en fracciones parciales (los pasos pueden verse »):

$$\frac{11 x^{2}}{2} + 17 {\color{red}{\int{\frac{1}{2 x^{2} + 7 x - 4} d x}}} = \frac{11 x^{2}}{2} + 17 {\color{red}{\int{\left(\frac{2}{9 \left(2 x - 1\right)} - \frac{1}{9 \left(x + 4\right)}\right)d x}}}$$

Integra término a término:

$$\frac{11 x^{2}}{2} + 17 {\color{red}{\int{\left(\frac{2}{9 \left(2 x - 1\right)} - \frac{1}{9 \left(x + 4\right)}\right)d x}}} = \frac{11 x^{2}}{2} + 17 {\color{red}{\left(- \int{\frac{1}{9 \left(x + 4\right)} d x} + \int{\frac{2}{9 \left(2 x - 1\right)} d x}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{9}$$$ y $$$f{\left(x \right)} = \frac{1}{x + 4}$$$:

$$\frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - 17 {\color{red}{\int{\frac{1}{9 \left(x + 4\right)} d x}}} = \frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - 17 {\color{red}{\left(\frac{\int{\frac{1}{x + 4} d x}}{9}\right)}}$$

Sea $$$u=x + 4$$$.

Entonces $$$du=\left(x + 4\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.

La integral se convierte en

$$\frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - \frac{17 {\color{red}{\int{\frac{1}{x + 4} d x}}}}{9} = \frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - \frac{17 {\color{red}{\int{\frac{1}{u} d u}}}}{9}$$

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

$$\frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - \frac{17 {\color{red}{\int{\frac{1}{u} d u}}}}{9} = \frac{11 x^{2}}{2} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} - \frac{17 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{9}$$

Recordemos que $$$u=x + 4$$$:

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{9} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{{\color{red}{\left(x + 4\right)}}}\right| \right)}}{9} + 17 \int{\frac{2}{9 \left(2 x - 1\right)} d x}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{2}{9}$$$ y $$$f{\left(x \right)} = \frac{1}{2 x - 1}$$$:

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + 17 {\color{red}{\int{\frac{2}{9 \left(2 x - 1\right)} d x}}} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + 17 {\color{red}{\left(\frac{2 \int{\frac{1}{2 x - 1} d x}}{9}\right)}}$$

Sea $$$u=2 x - 1$$$.

Entonces $$$du=\left(2 x - 1\right)^{\prime }dx = 2 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{2}$$$.

Por lo tanto,

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{34 {\color{red}{\int{\frac{1}{2 x - 1} d x}}}}{9} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{34 {\color{red}{\int{\frac{1}{2 u} d u}}}}{9}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{34 {\color{red}{\int{\frac{1}{2 u} d u}}}}{9} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{34 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}}{9}$$

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

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 {\color{red}{\int{\frac{1}{u} d u}}}}{9} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{9}$$

Recordemos que $$$u=2 x - 1$$$:

$$\frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{9} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 \ln{\left(\left|{{\color{red}{\left(2 x - 1\right)}}}\right| \right)}}{9}$$

Por lo tanto,

$$\int{\left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)d x} = \frac{11 x^{2}}{2} - \frac{17 \ln{\left(\left|{x + 4}\right| \right)}}{9} + \frac{17 \ln{\left(\left|{2 x - 1}\right| \right)}}{9}$$

Simplificar:

$$\int{\left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)d x} = \frac{99 x^{2} - 34 \ln{\left(\left|{x + 4}\right| \right)} + 34 \ln{\left(\left|{2 x - 1}\right| \right)}}{18}$$

Añade la constante de integración:

$$\int{\left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)d x} = \frac{99 x^{2} - 34 \ln{\left(\left|{x + 4}\right| \right)} + 34 \ln{\left(\left|{2 x - 1}\right| \right)}}{18}+C$$

$$$\int \left(11 x + \frac{17}{2 x^{2} + 7 x - 4}\right)\, dx = \frac{99 x^{2} - 34 \ln\left(\left|{x + 4}\right|\right) + 34 \ln\left(\left|{2 x - 1}\right|\right)}{18} + C$$$A