What is the integral of 1/(x^3-16x) ?

The integral of 1/(x^3-16x) is -1/16 ln|x|+1/32 ln|x+4|+1/32 ln|x-4|+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^3-16x)

\int \frac{1}{x ^{3} - 16 x} d x

- \frac{1}{16} ln ∣ x ∣ + \frac{1}{32} ln ∣ x + 4 ∣ + \frac{1}{32} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{1}{x ^{3} - 16 x} d x

Factor

x^{3} - 16 x : - (- x^{3} + 16 x)

= \int \frac{1}{- ( - x ^{3} + 16 x )} d x

Take the constant out:

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

= - \int \frac{1}{- x ^{3} + 16 x} d x

Take the partial fraction of

\frac{1}{- x ^{3} + 16 x} : \frac{1}{16 x} - \frac{1}{32 ( x + 4 )} - \frac{1}{32 ( x - 4 )}

= - \int \frac{1}{16 x} - \frac{1}{32 ( x + 4 )} - \frac{1}{32 ( x - 4 )} 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}{16 x} d x - \int \frac{1}{32 ( x + 4 )} d x - \int \frac{1}{32 ( x - 4 )} d x)

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

\int \frac{1}{32 ( x + 4 )} d x = \frac{1}{32} ln ∣ x + 4 ∣

\int \frac{1}{32 ( x - 4 )} d x = \frac{1}{32} ln ∣ x - 4 ∣

= - (\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln ∣ x + 4 ∣ - \frac{1}{32} ln ∣ x - 4 ∣)

Simplify

= - \frac{1}{16} ln ∣ x ∣ + \frac{1}{32} ln ∣ x + 4 ∣ + \frac{1}{32} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{16} ln ∣ x ∣ + \frac{1}{32} ln ∣ x + 4 ∣ + \frac{1}{32} ln ∣ x - 4 ∣ + C

\int \frac{1}{9 e ^{v} - 1} d v

\int 7 ln (2 x) d x

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

\int \frac{6 y - 7}{( 2 y + 1 ) ( 4 y ^{2} + 1 )} d y

\int \frac{7 x ^{2}}{( 144 + x ^{2} ) ^{2}} d x

What is the integral of 1/(x^3-16x) ?
The integral of 1/(x^3-16x) is -1/16 ln|x|+1/32 ln|x+4|+1/32 ln|x-4|+C