Integral of $$$\frac{1}{x^{2} - 78 x}$$$
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Find $$$\int \frac{1}{x^{2} - 78 x}\, dx$$$.
Solution
Perform partial fraction decomposition (steps can be seen »):
$${\color{red}{\int{\frac{1}{x^{2} - 78 x} d x}}} = {\color{red}{\int{\left(\frac{1}{78 \left(x - 78\right)} - \frac{1}{78 x}\right)d x}}}$$
Integrate term by term:
$${\color{red}{\int{\left(\frac{1}{78 \left(x - 78\right)} - \frac{1}{78 x}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{78 x} d x} + \int{\frac{1}{78 \left(x - 78\right)} d x}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{78}$$$ and $$$f{\left(x \right)} = \frac{1}{x}$$$:
$$\int{\frac{1}{78 \left(x - 78\right)} d x} - {\color{red}{\int{\frac{1}{78 x} d x}}} = \int{\frac{1}{78 \left(x - 78\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x} d x}}{78}\right)}}$$
The integral of $$$\frac{1}{x}$$$ is $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:
$$\int{\frac{1}{78 \left(x - 78\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x} d x}}}}{78} = \int{\frac{1}{78 \left(x - 78\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{78}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{78}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 78}$$$:
$$- \frac{\ln{\left(\left|{x}\right| \right)}}{78} + {\color{red}{\int{\frac{1}{78 \left(x - 78\right)} d x}}} = - \frac{\ln{\left(\left|{x}\right| \right)}}{78} + {\color{red}{\left(\frac{\int{\frac{1}{x - 78} d x}}{78}\right)}}$$
Let $$$u=x - 78$$$.
Then $$$du=\left(x - 78\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.
The integral can be rewritten as
$$- \frac{\ln{\left(\left|{x}\right| \right)}}{78} + \frac{{\color{red}{\int{\frac{1}{x - 78} d x}}}}{78} = - \frac{\ln{\left(\left|{x}\right| \right)}}{78} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{78}$$
The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$- \frac{\ln{\left(\left|{x}\right| \right)}}{78} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{78} = - \frac{\ln{\left(\left|{x}\right| \right)}}{78} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{78}$$
Recall that $$$u=x - 78$$$:
$$- \frac{\ln{\left(\left|{x}\right| \right)}}{78} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{78} = - \frac{\ln{\left(\left|{x}\right| \right)}}{78} + \frac{\ln{\left(\left|{{\color{red}{\left(x - 78\right)}}}\right| \right)}}{78}$$
Therefore,
$$\int{\frac{1}{x^{2} - 78 x} d x} = - \frac{\ln{\left(\left|{x}\right| \right)}}{78} + \frac{\ln{\left(\left|{x - 78}\right| \right)}}{78}$$
Simplify:
$$\int{\frac{1}{x^{2} - 78 x} d x} = \frac{- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 78}\right| \right)}}{78}$$
Add the constant of integration:
$$\int{\frac{1}{x^{2} - 78 x} d x} = \frac{- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 78}\right| \right)}}{78}+C$$
Answer
$$$\int \frac{1}{x^{2} - 78 x}\, dx = \frac{- \ln\left(\left|{x}\right|\right) + \ln\left(\left|{x - 78}\right|\right)}{78} + C$$$A