What is the integral of-(6x^3+8x)/(x^4-16) ?

The integral of-(6x^3+8x)/(x^4-16) is -3/2 ln|x^4-16|+1/2 ln|(x^2)/4+1|-1/2 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

Popular >

integral of-(6x^3+8x)/(x^4-16)

\int - \frac{6 x ^{3} + 8 x}{x ^{4} - 16} d x

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

Solution steps

\int - \frac{6 x ^{3} + 8 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{6 x ^{3} + 8 x}{x ^{4} - 16} d x

Expand

\frac{6 x ^{3} + 8 x}{x ^{4} - 16} : \frac{6 x ^{3}}{x ^{4} - 16} + \frac{8 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{6 x ^{3}}{x ^{4} - 16} d x + \int \frac{8 x}{x ^{4} - 16} d x)

\int \frac{6 x ^{3}}{x ^{4} - 16} d x = \frac{3}{2} ln x^{4} - 16

\int \frac{8 x}{x ^{4} - 16} d x = - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + \frac{1}{2} ln \frac{x ^{2}}{4} - 1

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

Simplify

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

Add a constant to the solution

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

\frac{d y}{d t} = 0.1 ln (\frac{4000}{y}) y

ln (7 x)

f (x) = lo g_{5} (2 x^{2} + 7), at x = 3

\frac{d}{d x} (sin (x) + 2 sin^{2} (x))

1 0^{x}, 1

What is the integral of-(6x^3+8x)/(x^4-16) ?
The integral of-(6x^3+8x)/(x^4-16) is -3/2 ln|x^4-16|+1/2 ln|(x^2)/4+1|-1/2 ln|(x^2)/4-1|+C