What is the integral of ((7x+1))/(x^2-7x+6) ?

The integral of ((7x+1))/(x^2-7x+6) is

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integral of ((7x+1))/(x^2-7x+6)

\int \frac{( 7 x + 1 )}{x ^{2} - 7 x + 6} d x

14 (\frac{1}{4} ln 4 x^{2} - 28 x + 24 - \frac{7}{20} (ln - \frac{2}{5} + \frac{2}{5} x - ln - \frac{12}{5} + \frac{2}{5} x)) - \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} ln ∣ x - 6 ∣ + C

Solution steps

\int \frac{7 x + 1}{x ^{2} - 7 x + 6} d x

Expand

\frac{7 x + 1}{x ^{2} - 7 x + 6} : \frac{7 x}{x ^{2} - 7 x + 6} + \frac{1}{x ^{2} - 7 x + 6}

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{7 x}{x ^{2} - 7 x + 6} d x + \int \frac{1}{x ^{2} - 7 x + 6} d x

\int \frac{7 x}{x ^{2} - 7 x + 6} d x = 14 (\frac{1}{4} ln 4 x^{2} - 28 x + 24 - \frac{7}{20} (ln - \frac{2}{5} + \frac{2}{5} x - ln - \frac{12}{5} + \frac{2}{5} x))

\int \frac{1}{x ^{2} - 7 x + 6} d x = - \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} ln ∣ x - 6 ∣

= 14 (\frac{1}{4} ln 4 x^{2} - 28 x + 24 - \frac{7}{20} (ln - \frac{2}{5} + \frac{2}{5} x - ln - \frac{12}{5} + \frac{2}{5} x)) - \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} ln ∣ x - 6 ∣

Add a constant to the solution

= 14 (\frac{1}{4} ln 4 x^{2} - 28 x + 24 - \frac{7}{20} (ln - \frac{2}{5} + \frac{2}{5} x - ln - \frac{12}{5} + \frac{2}{5} x)) - \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} ln ∣ x - 6 ∣ + C

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What is the integral of ((7x+1))/(x^2-7x+6) ?
The integral of ((7x+1))/(x^2-7x+6) is