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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x}{x ^{2} - x - 2} d x

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

s \to 0 lim (\frac{1}{s ^{2} 0.1 s + 1})

\int 4 cos^{2} (t) d t

x \to 0 lim (\frac{sin ^{2} ( 6 x )}{x})

x \to \infty lim (2 x^{x})

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 9, y^{^{'}} (0) = - 6

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