What is the integral of (x^2-4)/(x(x^2-1)(x^2+1)) ?

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

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integral of (x^2-4)/(x(x^2-1)(x^2+1))

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

4 ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 1 ∣ - \frac{3}{4} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{4}{x} - \frac{3}{4 ( x + 1 )} - \frac{3}{4 ( x - 1 )} - \frac{5 x}{2 ( x ^{2} + 1 )} 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{4}{x} d x - \int \frac{3}{4 ( x + 1 )} d x - \int \frac{3}{4 ( x - 1 )} d x - \int \frac{5 x}{2 ( x ^{2} + 1 )} d x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int \frac{3}{4 ( x + 1 )} d x = \frac{3}{4} ln ∣ x + 1 ∣

\int \frac{3}{4 ( x - 1 )} d x = \frac{3}{4} ln ∣ x - 1 ∣

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

= 4 ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 1 ∣ - \frac{3}{4} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1

Add a constant to the solution

= 4 ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 1 ∣ - \frac{3}{4} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + C

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