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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

= \int \frac{2}{x} + \frac{- x - 1}{x ^{2} + 3} 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} + 3} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

Add a constant to the solution

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

y^{^{''}} - (y^{^{'}})^{2} = 0

\frac{a x + b}{c x + d}

\frac{d}{d x} (3 x^{6})

x \to 0 - lim (\frac{23}{tan ( x )})

f (x) = \frac{6}{x + 1}

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