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

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

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

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

- 3 ln ∣ x ∣ + ln ∣ x + 1 ∣ + 3 ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 2 x + 3}{x ^{3} - x} : - \frac{3}{x} + \frac{1}{x + 1} + \frac{3}{x - 1}

= \int - \frac{3}{x} + \frac{1}{x + 1} + \frac{3}{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{3}{x} d x + \int \frac{1}{x + 1} d x + \int \frac{3}{x - 1} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

= - 3 ln ∣ x ∣ + ln ∣ x + 1 ∣ + 3 ln ∣ x - 1 ∣

Add a constant to the solution

= - 3 ln ∣ x ∣ + ln ∣ x + 1 ∣ + 3 ln ∣ x - 1 ∣ + C

\int_{2}^{3} \frac{1}{x - 1} d x

x \to - 4 lim (6 x + 11)

\frac{d}{d x} (x^{2} \cdot tan (\frac{1}{x}))

x \to 0 lim (\frac{1}{cos ( 2 x )})

\int_{0}^{L} 0 d x

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