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

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

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

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

ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

= ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣

Add a constant to the solution

= ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣ + C

\frac{x + 1}{e ^{x}}

e^{- 2 x} + sin (3 x) + 3 x^{2} - 2 x + 5, 0

f (x) = \frac{6}{( 1 + e ^{- x} )}, at x = 0

\frac{d y}{d x} = \frac{y}{x} + 4 x + 1

\frac{\partial}{\partial x} (t e^{x t})

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