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

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 x} = 6 x (y - 1)^{\frac{2}{3}}

\int \frac{x ^{3} - 6 x - 20}{x + 5} d x

t sin (π) t

x \to 3 lim (x^{2} - 5)

y^{^{'}} = \frac{( x + 3 y )}{2 x}

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