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

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

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integral of (6x)/(x^3-8)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{x}{x ^{3} - 8} d x

Take the partial fraction of

\frac{x}{x ^{3} - 8} : \frac{1}{6 ( x - 2 )} + \frac{- x + 2}{6 ( x ^{2} + 2 x + 4 )}

= 6 \cdot \int \frac{1}{6 ( x - 2 )} + \frac{- x + 2}{6 ( x ^{2} + 2 x + 4 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int \frac{1}{6 ( x - 2 )} d x + \int \frac{- x + 2}{6 ( x ^{2} + 2 x + 4 )} d x)

\int \frac{1}{6 ( x - 2 )} d x = \frac{1}{6} ln ∣ x - 2 ∣

\int \frac{- x + 2}{6 ( x ^{2} + 2 x + 4 )} d x = \frac{1}{12 3} (- 3 ln x^{2} + 2 x + 4 + 6 arctan (\frac{1}{3} (x + 1)))

= 6 (\frac{1}{6} ln ∣ x - 2 ∣ + \frac{1}{12 3} (- 3 ln x^{2} + 2 x + 4 + 6 arctan (\frac{1}{3} (x + 1))))

Simplify

6 (\frac{1}{6} ln ∣ x - 2 ∣ + \frac{1}{12 3} (- 3 ln x^{2} + 2 x + 4 + 6 arctan (\frac{1}{3} (x + 1)))) : ln ∣ x - 2 ∣ - \frac{1}{2} ln x^{2} + 2 x + 4 + 3 arctan (\frac{1}{3} (x + 1))

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

Add a constant to the solution

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

\int sec^{2} (π x) d x

\int_{- 1}^{1} t (1 - t)^{2} d t

x \to 0 lim (x^{3} ln (x^{2} + 1))

n = 1 \sum \infty \frac{( - 3 ) ^{n - 1}}{8 ^{n}}

\frac{\partial}{\partial x} (4 (x - 1)^{2})

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