What is the integral of (3x-1)/(x^2-5x) ?

The integral of (3x-1)/(x^2-5x) is 1/5 ln|x|+14/5 ln|x-5|+C

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integral of (3x-1)/(x^2-5x)

\int \frac{3 x - 1}{x ^{2} - 5 x} d x

\frac{1}{5} ln ∣ x ∣ + \frac{14}{5} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{3 x - 1}{x ^{2} - 5 x} d x

Expand

\frac{3 x - 1}{x ^{2} - 5 x} : \frac{3 x}{x ^{2} - 5 x} - \frac{1}{x ^{2} - 5 x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3 x}{x ^{2} - 5 x} d x - \int \frac{1}{x ^{2} - 5 x} d x

\int \frac{3 x}{x ^{2} - 5 x} d x = 3 ln ∣ x - 5 ∣

\int \frac{1}{x ^{2} - 5 x} d x = - \frac{1}{5} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 5 ∣

= 3 ln ∣ x - 5 ∣ - (- \frac{1}{5} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 5 ∣)

Simplify

3 ln ∣ x - 5 ∣ - (- \frac{1}{5} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 5 ∣) : \frac{1}{5} ln ∣ x ∣ + \frac{14}{5} ln ∣ x - 5 ∣

= \frac{1}{5} ln ∣ x ∣ + \frac{14}{5} ln ∣ x - 5 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ x ∣ + \frac{14}{5} ln ∣ x - 5 ∣ + C

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What is the integral of (3x-1)/(x^2-5x) ?
The integral of (3x-1)/(x^2-5x) is 1/5 ln|x|+14/5 ln|x-5|+C