Integral of $$$- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}$$$

The calculator will find the integral/antiderivative of $$$- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}$$$, with steps shown.

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Find $$$\int \left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)\, dx$$$.

Solution

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-24$$$ and $$$f{\left(x \right)} = \frac{x}{\left(x - 5\right) \left(x - 3\right)}$$$:

$${\color{red}{\int{\left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)d x}}} = {\color{red}{\left(- 24 \int{\frac{x}{\left(x - 5\right) \left(x - 3\right)} d x}\right)}}$$

Perform partial fraction decomposition (steps can be seen »):

$$- 24 {\color{red}{\int{\frac{x}{\left(x - 5\right) \left(x - 3\right)} d x}}} = - 24 {\color{red}{\int{\left(- \frac{3}{2 \left(x - 3\right)} + \frac{5}{2 \left(x - 5\right)}\right)d x}}}$$

Integrate term by term:

$$- 24 {\color{red}{\int{\left(- \frac{3}{2 \left(x - 3\right)} + \frac{5}{2 \left(x - 5\right)}\right)d x}}} = - 24 {\color{red}{\left(\int{\frac{5}{2 \left(x - 5\right)} d x} - \int{\frac{3}{2 \left(x - 3\right)} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{3}{2}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 3}$$$:

$$- 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 24 {\color{red}{\int{\frac{3}{2 \left(x - 3\right)} d x}}} = - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 24 {\color{red}{\left(\frac{3 \int{\frac{1}{x - 3} d x}}{2}\right)}}$$

Let $$$u=x - 3$$$.

Then $$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral becomes

$$- 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\int{\frac{1}{x - 3} d x}}} = - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\int{\frac{1}{u} d u}}}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\int{\frac{1}{u} d u}}} = - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recall that $$$u=x - 3$$$:

$$36 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} = 36 \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)} - 24 \int{\frac{5}{2 \left(x - 5\right)} d x}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{5}{2}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 5}$$$:

$$36 \ln{\left(\left|{x - 3}\right| \right)} - 24 {\color{red}{\int{\frac{5}{2 \left(x - 5\right)} d x}}} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 24 {\color{red}{\left(\frac{5 \int{\frac{1}{x - 5} d x}}{2}\right)}}$$

Let $$$u=x - 5$$$.

Then $$$du=\left(x - 5\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral becomes

$$36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\int{\frac{1}{x - 5} d x}}} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\int{\frac{1}{u} d u}}}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\int{\frac{1}{u} d u}}} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recall that $$$u=x - 5$$$:

$$36 \ln{\left(\left|{x - 3}\right| \right)} - 60 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 60 \ln{\left(\left|{{\color{red}{\left(x - 5\right)}}}\right| \right)}$$

Therefore,

$$\int{\left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)d x} = - 60 \ln{\left(\left|{x - 5}\right| \right)} + 36 \ln{\left(\left|{x - 3}\right| \right)}$$

Add the constant of integration:

$$\int{\left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)d x} = - 60 \ln{\left(\left|{x - 5}\right| \right)} + 36 \ln{\left(\left|{x - 3}\right| \right)}+C$$

Answer

$$$\int \left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)\, dx = \left(- 60 \ln\left(\left|{x - 5}\right|\right) + 36 \ln\left(\left|{x - 3}\right|\right)\right) + C$$$A


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