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

The integral of 1/((x-1)(x+2)(x+3)) is 1/12 ln|x-1|-1/3 ln|x+2|+1/4 ln|x+3|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{12} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 1 ) ( x + 2 ) ( x + 3 )} : \frac{1}{12 ( x - 1 )} - \frac{1}{3 ( x + 2 )} + \frac{1}{4 ( x + 3 )}

= \int \frac{1}{12 ( x - 1 )} - \frac{1}{3 ( x + 2 )} + \frac{1}{4 ( x + 3 )} 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

= \int \frac{1}{12 ( x - 1 )} d x - \int \frac{1}{3 ( x + 2 )} d x + \int \frac{1}{4 ( x + 3 )} d x

\int \frac{1}{12 ( x - 1 )} d x = \frac{1}{12} ln ∣ x - 1 ∣

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

\int \frac{1}{4 ( x + 3 )} d x = \frac{1}{4} ln ∣ x + 3 ∣

= \frac{1}{12} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{1}{12} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x + 3 ∣ + C

\frac{\partial}{\partial x} (sin (π (4 x - 4 y)))

\frac{2}{x}

(\frac{x ^{3}}{x ^{8} + 7})^{2}

h \to 0 lim (h^{4} cos (\frac{3}{h}))

\frac{\partial}{\partial x} (8 e^{x^{8} y})

What is the integral of 1/((x-1)(x+2)(x+3)) ?
The integral of 1/((x-1)(x+2)(x+3)) is 1/12 ln|x-1|-1/3 ln|x+2|+1/4 ln|x+3|+C