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

The integral of (x^2+2x-6)/(x^3-x) is 6ln|x|-7/2 ln|x+1|-3/2 ln|x-1|+C

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

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

6 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 1 ∣ - \frac{3}{2} ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 2 x - 6}{x ^{3} - x} : \frac{6}{x} - \frac{7}{2 ( x + 1 )} - \frac{3}{2 ( x - 1 )}

= \int \frac{6}{x} - \frac{7}{2 ( x + 1 )} - \frac{3}{2 ( 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{7}{2 ( x + 1 )} d x - \int \frac{3}{2 ( x - 1 )} d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

\int \frac{7}{2 ( x + 1 )} d x = \frac{7}{2} ln ∣ x + 1 ∣

\int \frac{3}{2 ( x - 1 )} d x = \frac{3}{2} ln ∣ x - 1 ∣

= 6 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 1 ∣ - \frac{3}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= 6 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 1 ∣ - \frac{3}{2} ln ∣ x - 1 ∣ + C

\frac{d}{d x} (e^{x \cdot l n (x)})

\frac{d}{d x} (\frac{1}{x ln ^{2} ( x )})

y = tan (12 x^{2}), at x = π

(x^{2})^{2}

sec^{2} (0)

What is the integral of (x^2+2x-6)/(x^3-x) ?
The integral of (x^2+2x-6)/(x^3-x) is 6ln|x|-7/2 ln|x+1|-3/2 ln|x-1|+C