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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{x} - \frac{4}{9 x ^{2} + 1} d x

\int sin^{2} (\frac{u}{2}) d u

\int sin^{7} (\frac{x}{4}) cos (\frac{x}{4}) d x

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

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