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

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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

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

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\frac{d}{d x} (e x)

\frac{\partial}{\partial v} (\frac{u}{u + v})

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