What is the integral of (18)/((x+3)(x^2-9)) ?

The integral of (18)/((x+3)(x^2-9)) is 18(-1/36 ln|x+3|+1/(6(x+3))+1/36 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 (18)/((x+3)(x^2-9))

\int \frac{18}{( x + 3 ) ( x ^{2} - 9 )} d x

18 (- \frac{1}{36} ln ∣ x + 3 ∣ + \frac{1}{6 ( x + 3 )} + \frac{1}{36} ln ∣ x - 3 ∣) + C

Solution steps

\int \frac{18}{( x + 3 ) ( x ^{2} - 9 )} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{( x + 3 ) ( x ^{2} - 9 )} d x

Take the partial fraction of

\frac{1}{( x + 3 ) ( x ^{2} - 9 )} : - \frac{1}{36 ( x + 3 )} - \frac{1}{6 ( x + 3 ) ^{2}} + \frac{1}{36 ( x - 3 )}

= 18 \cdot \int - \frac{1}{36 ( x + 3 )} - \frac{1}{6 ( x + 3 ) ^{2}} + \frac{1}{36 ( 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

= 18 (- \int \frac{1}{36 ( x + 3 )} d x - \int \frac{1}{6 ( x + 3 ) ^{2}} d x + \int \frac{1}{36 ( x - 3 )} d x)

\int \frac{1}{36 ( x + 3 )} d x = \frac{1}{36} ln ∣ x + 3 ∣

\int \frac{1}{6 ( x + 3 ) ^{2}} d x = - \frac{1}{6 ( x + 3 )}

\int \frac{1}{36 ( x - 3 )} d x = \frac{1}{36} ln ∣ x - 3 ∣

= 18 (- \frac{1}{36} ln ∣ x + 3 ∣ - (- \frac{1}{6 ( x + 3 )}) + \frac{1}{36} ln ∣ x - 3 ∣)

Simplify

= 18 (- \frac{1}{36} ln ∣ x + 3 ∣ + \frac{1}{6 ( x + 3 )} + \frac{1}{36} ln ∣ x - 3 ∣)

Add a constant to the solution

= 18 (- \frac{1}{36} ln ∣ x + 3 ∣ + \frac{1}{6 ( x + 3 )} + \frac{1}{36} ln ∣ x - 3 ∣) + C

y^{^{''}} + 2 y^{^{'}} + y = \frac{- 4 e ^{- t}}{t ^{2} + 1}

\frac{\partial}{\partial x} (\frac{y ^{2} - x ^{2}}{x y})

n = 1 \sum \infty \frac{4}{7 n + 2}

\frac{d}{d x} (x^{5} e^{- 3 x})

\frac{d}{d t} (7 e^{t} sin (t))

What is the integral of (18)/((x+3)(x^2-9)) ?
The integral of (18)/((x+3)(x^2-9)) is 18(-1/36 ln|x+3|+1/(6(x+3))+1/36 ln|x-3|)+C