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

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

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integral of (x^2-3x-1)/(x^3+x^2-2x)

\int \frac{x ^{2} - 3 x - 1}{x ^{3} + x ^{2} - 2 x} d x

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

Solution steps

\int \frac{x ^{2} - 3 x - 1}{x ^{3} + x ^{2} - 2 x} d x

Take the partial fraction of

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

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

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\int_{0}^{8} y^{2} (8 - y) d y

4 x - 5 x^{\frac{3}{4}}

\frac{d}{d x} (e^{- x} - x)

\int \frac{1}{( x + 1 ) ln ( x + 1 )} d x

y = 3 - 2 x + 4 x^{3}, at x = - 1

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