What is the integral of 1/(x^3+8x) ?

The integral of 1/(x^3+8x) is 1/8 ln|x|-1/16 ln|x^2+8|+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^3+8x)

\int \frac{1}{x ^{3} + 8 x} d x

\frac{1}{8} ln ∣ x ∣ - \frac{1}{16} ln x^{2} + 8 + C

Solution steps

\int \frac{1}{x ^{3} + 8 x} d x

Take the partial fraction of

\frac{1}{x ^{3} + 8 x} : \frac{1}{8 x} - \frac{x}{8 ( x ^{2} + 8 )}

= \int \frac{1}{8 x} - \frac{x}{8 ( x ^{2} + 8 )} 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}{8 x} d x - \int \frac{x}{8 ( x ^{2} + 8 )} d x

\int \frac{1}{8 x} d x = \frac{1}{8} ln ∣ x ∣

\int \frac{x}{8 ( x ^{2} + 8 )} d x = \frac{1}{16} ln x^{2} + 8

= \frac{1}{8} ln ∣ x ∣ - \frac{1}{16} ln x^{2} + 8

Add a constant to the solution

= \frac{1}{8} ln ∣ x ∣ - \frac{1}{16} ln x^{2} + 8 + C

f (x) = 4 cos (2 x)

n = 0 \sum \infty \frac{4}{n}

y = tan^{2} (x), (\frac{π}{4}, 1)

\frac{d}{d x} (e^{\frac{1}{2} x^{2}})

\frac{d}{d x} (2^{3 x + 1} ln (5 x - 11))

What is the integral of 1/(x^3+8x) ?
The integral of 1/(x^3+8x) is 1/8 ln|x|-1/16 ln|x^2+8|+C