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

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

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

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

- 4 ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + ln ∣ x + 3 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \int \frac{1}{x ^{3} + 2 x ^{2} - 3 x} d x

Take the partial fraction of

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

= 12 \cdot \int - \frac{1}{3 x} + \frac{1}{4 ( x - 1 )} + \frac{1}{12 ( 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

= 12 (- \int \frac{1}{3 x} d x + \int \frac{1}{4 ( x - 1 )} d x + \int \frac{1}{12 ( x + 3 )} d x)

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{1}{12 ( x + 3 )} d x = \frac{1}{12} ln ∣ x + 3 ∣

= 12 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣)

Simplify

12 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣) : - 4 ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + ln ∣ x + 3 ∣

= - 4 ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + ln ∣ x + 3 ∣

Add a constant to the solution

= - 4 ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + ln ∣ x + 3 ∣ + C

y^{^{''}} - y^{^{'}} - 4 y = 0

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

x \to \infty lim (\frac{1}{x} + x)

4 e^{t}

x \to 4 lim (\frac{2}{( x + 4 ) ^{3}})

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