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

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

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

\int \frac{24 x - 2 x ^{3}}{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 + C

Solution steps

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

Expand

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

\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)

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

Add a constant to the solution

= - 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 + C

\frac{d}{d x} (\frac{8}{x} - \frac{4}{x})

\frac{\partial}{\partial y} (yz - ln (x + z))

x \to 2 lim (3 x^{2} + 1)

y = 300 (1.056)^{x}

\frac{\partial}{\partial x} (3 x^{2} + 2 x y^{2} + 1)

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