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

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

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

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

ln ∣ x ∣ - \frac{5}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{1}{x} - \frac{5}{4 ( x + 2 )} + \frac{1}{4 ( 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}{4 ( x + 2 )} d x + \int \frac{1}{4 ( x - 2 )} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{1}{4 ( x - 2 )} d x = \frac{1}{4} ln ∣ x - 2 ∣

= ln ∣ x ∣ - \frac{5}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣

Add a constant to the solution

= ln ∣ x ∣ - \frac{5}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣ + C

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