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

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

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

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

ln ∣ x + 1 ∣ + 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{3 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

= 3 \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 )}

= 3 \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

= 3 (\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 ∣

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

Simplify

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

= ln ∣ x + 1 ∣ + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ + 2 ln ∣ x - 2 ∣ + C

(x - 1) d y + (y - x^{2} + 2 x^{- 3}) d x = 0

x = y^{\frac{1}{2}}, x = 2 [0, 2]

f (x) = x^{2}, at x = 0.2

f (x) = \frac{7}{x ^{2}}

\int \frac{1}{3} sec (x) tan (x) d x

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