What is the integral of (3x-4)/(x^2-x-6) ?

The integral of (3x-4)/(x^2-x-6) is 2ln|x+2|+ln|x-3|+C

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integral of (3x-4)/(x^2-x-6)

\int \frac{3 x - 4}{x ^{2} - x - 6} d x

2 ln ∣ x + 2 ∣ + ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{3 x - 4}{x ^{2} - x - 6} d x

Take the partial fraction of

\frac{3 x - 4}{x ^{2} - x - 6} : \frac{2}{x + 2} + \frac{1}{x - 3}

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

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

\int \frac{1}{x - 3} d x = ln ∣ x - 3 ∣

= 2 ln ∣ x + 2 ∣ + ln ∣ x - 3 ∣

Add a constant to the solution

= 2 ln ∣ x + 2 ∣ + ln ∣ x - 3 ∣ + C

\frac{d}{d x} (6 x - 3 x)

x \to 5 - lim (c x^{2} + 2 x)

\int \frac{8 x ^{3} + 7}{( x + 1 ) ( 2 x + 1 ) ^{3}} d x

y = \frac{x ^{2}}{2} - 4 x + 6, y = \frac{x}{2} - 1

x^{\frac{1}{2}} + y^{\frac{1}{2}} = 9

What is the integral of (3x-4)/(x^2-x-6) ?
The integral of (3x-4)/(x^2-x-6) is 2ln|x+2|+ln|x-3|+C