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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/(x^3+9x)

\int \frac{2}{x ^{3} + 9 x} d x

2 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) + C

Solution steps

\int \frac{2}{x ^{3} + 9 x} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ^{3} + 9 x} d x

Take the partial fraction of

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

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

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

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

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

= 2 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9)

Add a constant to the solution

= 2 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) + C

\frac{d}{d x} ((tan (x))^{s i n (x)})

\frac{d}{d x} (- 2 e^{2 x})

x \to 6 lim (\frac{7}{x - 6})

\int_{0}^{4} 2 x^{2} - 7 x + 3 d x

(\frac{d y}{d x}) + \frac{2 x y}{x ^{2} + 1} = x

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