What is the integral of 9/((x-7)(x+2)) ?

The integral of 9/((x-7)(x+2)) is ln|x-7|-ln|x+2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9/((x-7)(x+2))

\int \frac{9}{( x - 7 ) ( x + 2 )} d x

ln ∣ x - 7 ∣ - ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{9}{( x - 7 ) ( x + 2 )} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{( x - 7 ) ( x + 2 )} d x

Take the partial fraction of

\frac{1}{( x - 7 ) ( x + 2 )} : \frac{1}{9 ( x - 7 )} - \frac{1}{9 ( x + 2 )}

= 9 \cdot \int \frac{1}{9 ( x - 7 )} - \frac{1}{9 ( 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

= 9 (\int \frac{1}{9 ( x - 7 )} d x - \int \frac{1}{9 ( x + 2 )} d x)

\int \frac{1}{9 ( x - 7 )} d x = \frac{1}{9} ln ∣ x - 7 ∣

\int \frac{1}{9 ( x + 2 )} d x = \frac{1}{9} ln ∣ x + 2 ∣

= 9 (\frac{1}{9} ln ∣ x - 7 ∣ - \frac{1}{9} ln ∣ x + 2 ∣)

Simplify

9 (\frac{1}{9} ln ∣ x - 7 ∣ - \frac{1}{9} ln ∣ x + 2 ∣) : ln ∣ x - 7 ∣ - ln ∣ x + 2 ∣

= ln ∣ x - 7 ∣ - ln ∣ x + 2 ∣

Add a constant to the solution

= ln ∣ x - 7 ∣ - ln ∣ x + 2 ∣ + C

\frac{d x}{d t} = x b

2 x^{3} e^{- x}

f (x) = 100 - x^{2}, at x = - 6

t sin (1.74 t)

\int - \frac{1}{3 ( s - 1 ) ^{3}} d s

What is the integral of 9/((x-7)(x+2)) ?
The integral of 9/((x-7)(x+2)) is ln|x-7|-ln|x+2|+C