What is the integral of ((x+3))/(6x-x^2) ?

The integral of ((x+3))/(6x-x^2) is

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integral of ((x+3))/(6x-x^2)

\int \frac{( x + 3 )}{6 x - x ^{2}} d x

\frac{1}{6} (x + 3) (ln (\frac{1}{3} ∣ x ∣) - ln \frac{1}{3} x - 2) - \frac{1}{6} x ln (\frac{1}{3} ∣ x ∣) + \frac{1}{6} x ln \frac{1}{3} x - 2 - ln ∣ - x + 6 ∣ + C

Solution steps

\int \frac{x + 3}{6 x - x ^{2}} d x

Apply Integration By Parts

= \frac{1}{6} (x + 3) (ln \frac{1}{3} (x - 3) + 1 - ln \frac{1}{3} (x - 3) - 1) - \int \frac{1}{6} (ln \frac{1}{3} (x - 3) + 1 - ln \frac{1}{3} (x - 3) - 1) d x

\int \frac{1}{6} (ln \frac{1}{3} (x - 3) + 1 - ln \frac{1}{3} (x - 3) - 1) d x = \frac{1}{6} x ln (\frac{1}{3} ∣ x ∣) - \frac{1}{6} x ln \frac{1}{3} x - 2 + ln ∣ - x + 6 ∣

= \frac{1}{6} (x + 3) (ln \frac{1}{3} (x - 3) + 1 - ln \frac{1}{3} (x - 3) - 1) - (\frac{1}{6} x ln (\frac{1}{3} ∣ x ∣) - \frac{1}{6} x ln \frac{1}{3} x - 2 + ln ∣ - x + 6 ∣)

Simplify

\frac{1}{6} (x + 3) (ln \frac{1}{3} (x - 3) + 1 - ln \frac{1}{3} (x - 3) - 1) - (\frac{1}{6} x ln (\frac{1}{3} ∣ x ∣) - \frac{1}{6} x ln \frac{1}{3} x - 2 + ln ∣ - x + 6 ∣) : \frac{1}{6} (x + 3) (ln (\frac{1}{3} ∣ x ∣) - ln \frac{1}{3} x - 2) - \frac{1}{6} x ln (\frac{1}{3} ∣ x ∣) + \frac{1}{6} x ln \frac{1}{3} x - 2 - ln ∣ - x + 6 ∣

= \frac{1}{6} (x + 3) (ln (\frac{1}{3} ∣ x ∣) - ln \frac{1}{3} x - 2) - \frac{1}{6} x ln (\frac{1}{3} ∣ x ∣) + \frac{1}{6} x ln \frac{1}{3} x - 2 - ln ∣ - x + 6 ∣

Add a constant to the solution

= \frac{1}{6} (x + 3) (ln (\frac{1}{3} ∣ x ∣) - ln \frac{1}{3} x - 2) - \frac{1}{6} x ln (\frac{1}{3} ∣ x ∣) + \frac{1}{6} x ln \frac{1}{3} x - 2 - ln ∣ - x + 6 ∣ + C