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(1/2 ln|(x^2)/4+1|-1/2 ln|(x^2)/4-1|)+C

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

Solution steps

\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)

Add a constant to the solution

= \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) + C

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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(1/2 ln|(x^2)/4+1|-1/2 ln|(x^2)/4-1|)+C