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

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

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

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

ln ∣ x + 1 ∣ + 2 ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x + 1 ) ( 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

y^{^{''}} + 4 y^{^{'}} + 4 y = x e^{- 2 x}

n \to \infty lim (\frac{1}{5})

y^{^{'}} = \frac{2 y + x}{x}

n = 0 \sum \infty ln (2)

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