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

The integral of (x^4+2x^2+3)/(x^3-4x) is (x^2)/2-3/4 ln|x|+27/8 ln|x+2|+27/8 ln|x-2|+C

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

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

\frac{x ^{2}}{2} - \frac{3}{4} ln ∣ x ∣ + \frac{27}{8} ln ∣ x + 2 ∣ + \frac{27}{8} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4} + 2 x ^{2} + 3}{x ^{3} - 4 x} : x - \frac{3}{4 x} + \frac{27}{8 ( x + 2 )} + \frac{27}{8 ( x - 2 )}

= \int x - \frac{3}{4 x} + \frac{27}{8 ( x + 2 )} + \frac{27}{8 ( 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 x d x - \int \frac{3}{4 x} d x + \int \frac{27}{8 ( x + 2 )} d x + \int \frac{27}{8 ( x - 2 )} d x

\int x d x = \frac{x ^{2}}{2}

\int \frac{3}{4 x} d x = \frac{3}{4} ln ∣ x ∣

\int \frac{27}{8 ( x + 2 )} d x = \frac{27}{8} ln ∣ x + 2 ∣

\int \frac{27}{8 ( x - 2 )} d x = \frac{27}{8} ln ∣ x - 2 ∣

= \frac{x ^{2}}{2} - \frac{3}{4} ln ∣ x ∣ + \frac{27}{8} ln ∣ x + 2 ∣ + \frac{27}{8} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{3}{4} ln ∣ x ∣ + \frac{27}{8} ln ∣ x + 2 ∣ + \frac{27}{8} ln ∣ x - 2 ∣ + C

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What is the integral of (x^4+2x^2+3)/(x^3-4x) ?
The integral of (x^4+2x^2+3)/(x^3-4x) is (x^2)/2-3/4 ln|x|+27/8 ln|x+2|+27/8 ln|x-2|+C