What is the integral of (3x^2-2)/(x^2-4x-12) ?

The integral of (3x^2-2)/(x^2-4x-12) is 3x-5/4 ln|x+2|+53/4 ln|x-6|+C

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integral of (3x^2-2)/(x^2-4x-12)

\int \frac{3 x ^{2} - 2}{x ^{2} - 4 x - 12} d x

3 x - \frac{5}{4} ln ∣ x + 2 ∣ + \frac{53}{4} ln ∣ x - 6 ∣ + C

Solution steps

\int \frac{3 x ^{2} - 2}{x ^{2} - 4 x - 12} d x

Take the partial fraction of

\frac{3 x ^{2} - 2}{x ^{2} - 4 x - 12} : 3 - \frac{5}{4 ( x + 2 )} + \frac{53}{4 ( x - 6 )}

= \int 3 - \frac{5}{4 ( x + 2 )} + \frac{53}{4 ( x - 6 )} 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 3 d x - \int \frac{5}{4 ( x + 2 )} d x + \int \frac{53}{4 ( x - 6 )} d x

\int 3 d x = 3 x

\int \frac{5}{4 ( x + 2 )} d x = \frac{5}{4} ln ∣ x + 2 ∣

\int \frac{53}{4 ( x - 6 )} d x = \frac{53}{4} ln ∣ x - 6 ∣

= 3 x - \frac{5}{4} ln ∣ x + 2 ∣ + \frac{53}{4} ln ∣ x - 6 ∣

Add a constant to the solution

= 3 x - \frac{5}{4} ln ∣ x + 2 ∣ + \frac{53}{4} ln ∣ x - 6 ∣ + C

\frac{d}{d x} (arcsin (x^{100}))

\frac{20}{2 x ^{2} + 9 x + 30}

e^{3 t} sin (6 t)

y = x

\frac{\partial}{\partial w} (x e^{w})

What is the integral of (3x^2-2)/(x^2-4x-12) ?
The integral of (3x^2-2)/(x^2-4x-12) is 3x-5/4 ln|x+2|+53/4 ln|x-6|+C