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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

f (x) = (x^{3} - 7^{x} + 3), at x = 1

y^{^{''}} - 9 y^{^{'}} = 7 t

f (x) = \frac{tan ( x ^{2} )}{4}

y = \frac{x ^{8}}{f ( x )}

\int 5 cot^{5} (θ) sin^{4} (θ) d θ

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