What is the integral of (2x^3+x^2+6)/(x^5-x) ?

The integral of (2x^3+x^2+6)/(x^5-x) is -6ln|x|+5/4 ln|x^2+1|+arctan(x)+5/4 ln|x+1|+9/4 ln|x-1|+C

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integral of (2x^3+x^2+6)/(x^5-x)

\int \frac{2 x ^{3} + x ^{2} + 6}{x ^{5} - x} d x

- 6 ln ∣ x ∣ + \frac{5}{4} ln x^{2} + 1 + arctan (x) + \frac{5}{4} ln ∣ x + 1 ∣ + \frac{9}{4} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{2 x ^{3} + x ^{2} + 6}{x ^{5} - x} d x

Take the partial fraction of

\frac{2 x ^{3} + x ^{2} + 6}{x ^{5} - x} : - \frac{6}{x} + \frac{5 x + 2}{2 ( x ^{2} + 1 )} + \frac{5}{4 ( x + 1 )} + \frac{9}{4 ( x - 1 )}

= \int - \frac{6}{x} + \frac{5 x + 2}{2 ( x ^{2} + 1 )} + \frac{5}{4 ( x + 1 )} + \frac{9}{4 ( x - 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{6}{x} d x + \int \frac{5 x + 2}{2 ( x ^{2} + 1 )} d x + \int \frac{5}{4 ( x + 1 )} d x + \int \frac{9}{4 ( x - 1 )} d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

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

\int \frac{9}{4 ( x - 1 )} d x = \frac{9}{4} ln ∣ x - 1 ∣

= - 6 ln ∣ x ∣ + \frac{5}{4} ln x^{2} + 1 + arctan (x) + \frac{5}{4} ln ∣ x + 1 ∣ + \frac{9}{4} ln ∣ x - 1 ∣

Add a constant to the solution

= - 6 ln ∣ x ∣ + \frac{5}{4} ln x^{2} + 1 + arctan (x) + \frac{5}{4} ln ∣ x + 1 ∣ + \frac{9}{4} ln ∣ x - 1 ∣ + C

What is the integral of (2x^3+x^2+6)/(x^5-x) ?
The integral of (2x^3+x^2+6)/(x^5-x) is -6ln|x|+5/4 ln|x^2+1|+arctan(x)+5/4 ln|x+1|+9/4 ln|x-1|+C