What is the integral of (x^2-x+12)/(x^3+6x) ?

The integral of (x^2-x+12)/(x^3+6x) is 2ln|x|-1/2 ln|x^2+6|-1/(sqrt(6))arctan(x/(sqrt(6)))+C

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integral of (x^2-x+12)/(x^3+6x)

\int \frac{x ^{2} - x + 12}{x ^{3} + 6 x} d x

2 ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 6 - \frac{1}{6} arctan (\frac{x}{6}) + C

Solution steps

\int \frac{x ^{2} - x + 12}{x ^{3} + 6 x} d x

Take the partial fraction of

\frac{x ^{2} - x + 12}{x ^{3} + 6 x} : \frac{2}{x} + \frac{- x - 1}{x ^{2} + 6}

= \int \frac{2}{x} + \frac{- x - 1}{x ^{2} + 6} 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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (e^{3 x} sin (cy))

n = 0 \sum \infty \frac{3 ^{n}}{2 ^{n}}

f (x) = \frac{1}{2 - 3 x}

\frac{d}{d x} (ln (\frac{1 + x}{x ^{4}}))

n = 6 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

What is the integral of (x^2-x+12)/(x^3+6x) ?
The integral of (x^2-x+12)/(x^3+6x) is 2ln|x|-1/2 ln|x^2+6|-1/(sqrt(6))arctan(x/(sqrt(6)))+C