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

The integral of (x^2-x-3)/(x^2-3x) is (x^2)/(x-3)+ln|x-3|-9/(x-3)+ln|x|+C

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

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

\frac{x ^{2}}{x - 3} + ln ∣ x - 3 ∣ - \frac{9}{x - 3} + ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{x ^{2} - x - 3}{x ^{2} - 3 x} : \frac{x ^{2}}{x ^{2} - 3 x} - \frac{x}{x ^{2} - 3 x} - \frac{3}{x ^{2} - 3 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{x ^{2}}{x ^{2} - 3 x} d x - \int \frac{x}{x ^{2} - 3 x} d x - \int \frac{3}{x ^{2} - 3 x} d x

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

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

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

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

Simplify

\frac{x ^{2}}{x - 3} + 3 (ln ∣ x - 3 ∣ - \frac{3}{x - 3}) - ln ∣ x - 3 ∣ - (- ln ∣ x ∣ + ln ∣ x - 3 ∣) : \frac{x ^{2}}{x - 3} + ln ∣ x - 3 ∣ - \frac{9}{x - 3} + ln ∣ x ∣

= \frac{x ^{2}}{x - 3} + ln ∣ x - 3 ∣ - \frac{9}{x - 3} + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{x - 3} + ln ∣ x - 3 ∣ - \frac{9}{x - 3} + ln ∣ x ∣ + C

\frac{d}{d x} (arccos (e^{6 x}))

\int (6 t^{2} + 4 t) (t^{3} + t^{2} + 1)^{6} d t

x \to - 2 - lim (x + 2)

y^{^{''}} + 6 y^{^{'}} + 8 y = e^{5 x}, y (0) = 2

\frac{d y}{d x} + 7 y = 12

What is the integral of (x^2-x-3)/(x^2-3x) ?
The integral of (x^2-x-3)/(x^2-3x) is (x^2)/(x-3)+ln|x-3|-9/(x-3)+ln|x|+C