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

The integral of (x^2+4)/(3x^3-4x^2-4x) is -ln|x|+5/6 ln|3x+2|+1/2 ln|x-2|+C

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

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

- ln ∣ x ∣ + \frac{5}{6} ln ∣ 3 x + 2 ∣ + \frac{1}{2} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{5}{2 ( 3 x + 2 )} d x = \frac{5}{6} ln ∣ 3 x + 2 ∣

\int \frac{1}{2 ( x - 2 )} d x = \frac{1}{2} ln ∣ x - 2 ∣

= - ln ∣ x ∣ + \frac{5}{6} ln ∣ 3 x + 2 ∣ + \frac{1}{2} ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x ∣ + \frac{5}{6} ln ∣ 3 x + 2 ∣ + \frac{1}{2} ln ∣ x - 2 ∣ + C

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} - 3 y = 0

x \to 4 lim ([4 f (8) g (- 1)])

\int \frac{1}{y ^{2} - 2} d y

2 (t - 2)^{2} + 6 (t - 2) + 6

\int \frac{1}{4 x ^{2} - 16} d x

What is the integral of (x^2+4)/(3x^3-4x^2-4x) ?
The integral of (x^2+4)/(3x^3-4x^2-4x) is -ln|x|+5/6 ln|3x+2|+1/2 ln|x-2|+C