What is the integral of-(2x^3+24x)/(x^4-16) ?

The integral of-(2x^3+24x)/(x^4-16) is -1/2 ln|x^4-16|+3/2 ln|(x^2)/4+1|-3/2 ln|(x^2)/4-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of-(2x^3+24x)/(x^4-16)

\int - \frac{2 x ^{3} + 24 x}{x ^{4} - 16} d x

- \frac{1}{2} ln x^{4} - 16 + \frac{3}{2} ln \frac{x ^{2}}{4} + 1 - \frac{3}{2} ln \frac{x ^{2}}{4} - 1 + C

Solution steps

\int - \frac{2 x ^{3} + 24 x}{x ^{4} - 16} d x

Take the constant out:

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

= - \int \frac{2 x ^{3} + 24 x}{x ^{4} - 16} d x

Expand

\frac{2 x ^{3} + 24 x}{x ^{4} - 16} : \frac{2 x ^{3}}{x ^{4} - 16} + \frac{24 x}{x ^{4} - 16}

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{2 x ^{3}}{x ^{4} - 16} d x + \int \frac{24 x}{x ^{4} - 16} d x)

\int \frac{2 x ^{3}}{x ^{4} - 16} d x = \frac{1}{2} ln x^{4} - 16

\int \frac{24 x}{x ^{4} - 16} d x = - 3 (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - \frac{1}{2} ln \frac{x ^{2}}{4} - 1)

= - (\frac{1}{2} ln x^{4} - 16 - 3 (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - \frac{1}{2} ln \frac{x ^{2}}{4} - 1))

Simplify

- (\frac{1}{2} ln x^{4} - 16 - 3 (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - \frac{1}{2} ln \frac{x ^{2}}{4} - 1)) : - \frac{1}{2} ln x^{4} - 16 + \frac{3}{2} ln \frac{x ^{2}}{4} + 1 - \frac{3}{2} ln \frac{x ^{2}}{4} - 1

= - \frac{1}{2} ln x^{4} - 16 + \frac{3}{2} ln \frac{x ^{2}}{4} + 1 - \frac{3}{2} ln \frac{x ^{2}}{4} - 1

Add a constant to the solution

= - \frac{1}{2} ln x^{4} - 16 + \frac{3}{2} ln \frac{x ^{2}}{4} + 1 - \frac{3}{2} ln \frac{x ^{2}}{4} - 1 + C

What is the integral of-(2x^3+24x)/(x^4-16) ?
The integral of-(2x^3+24x)/(x^4-16) is -1/2 ln|x^4-16|+3/2 ln|(x^2)/4+1|-3/2 ln|(x^2)/4-1|+C