What is the integral of (4x^2-7x-12)/(x(x+2)(x-3)) ?

The integral of (4x^2-7x-12)/(x(x+2)(x-3)) is 2ln|x|+9/5 ln|x+2|+1/5 ln|x-3|+C

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integral of (4x^2-7x-12)/(x(x+2)(x-3))

\int \frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} d x

2 ln ∣ x ∣ + \frac{9}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} d x

Take the partial fraction of

\frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} : \frac{2}{x} + \frac{9}{5 ( x + 2 )} + \frac{1}{5 ( x - 3 )}

= \int \frac{2}{x} + \frac{9}{5 ( x + 2 )} + \frac{1}{5 ( 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} d x + \int \frac{9}{5 ( x + 2 )} d x + \int \frac{1}{5 ( x - 3 )} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{9}{5 ( x + 2 )} d x = \frac{9}{5} ln ∣ x + 2 ∣

\int \frac{1}{5 ( x - 3 )} d x = \frac{1}{5} ln ∣ x - 3 ∣

= 2 ln ∣ x ∣ + \frac{9}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣

Add a constant to the solution

= 2 ln ∣ x ∣ + \frac{9}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣ + C

\int \frac{4}{x ^{2} ( x ^{2} + 4 )} d x

x^{3}, 3

\int x^{10} e^{x^{11}} d x

\frac{d}{d x} (7 sin^{2} (x))

f (x) = 1 - 3 x, at x = - 1

What is the integral of (4x^2-7x-12)/(x(x+2)(x-3)) ?
The integral of (4x^2-7x-12)/(x(x+2)(x-3)) is 2ln|x|+9/5 ln|x+2|+1/5 ln|x-3|+C