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

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

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

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

- \frac{3}{2} ln ∣ x ∣ + \frac{5}{3} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{3}{2 x} + \frac{5}{3 ( x - 1 )} - \frac{1}{6 ( 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 \frac{3}{2 x} d x + \int \frac{5}{3 ( x - 1 )} d x - \int \frac{1}{6 ( x + 2 )} d x

\int \frac{3}{2 x} d x = \frac{3}{2} ln ∣ x ∣

\int \frac{5}{3 ( x - 1 )} d x = \frac{5}{3} ln ∣ x - 1 ∣

\int \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} ln ∣ x + 2 ∣

= - \frac{3}{2} ln ∣ x ∣ + \frac{5}{3} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 2 ∣

Add a constant to the solution

= - \frac{3}{2} ln ∣ x ∣ + \frac{5}{3} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 2 ∣ + C

n = 2 \sum \infty \frac{1}{n} (- 1)^{n}

y^{2} = 16 - x, (y + 2)^{2} = x + 4

\frac{e ^{s}}{7 s + 1}

f (x) = (x e^{2 x - 4}), x = 2

n = 1 \sum \infty \frac{n - 7}{n + 7}

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