What is the integral of ((x^2-x+24))/(x^3+6x) ?

The integral of ((x^2-x+24))/(x^3+6x) is 4ln|x|-3/2 ln|x^2+6|-1/(sqrt(6))arctan(x/(sqrt(6)))+C

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integral of ((x^2-x+24))/(x^3+6x)

\int \frac{( x ^{2} - x + 24 )}{x ^{3} + 6 x} d x

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

Solution steps

\int \frac{x ^{2} - x + 24}{x ^{3} + 6 x} d x

Take the partial fraction of

\frac{x ^{2} - x + 24}{x ^{3} + 6 x} : \frac{4}{x} + \frac{- 3 x - 1}{x ^{2} + 6}

= \int \frac{4}{x} + \frac{- 3 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{4}{x} d x + \int \frac{- 3 x - 1}{x ^{2} + 6} d x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

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

Add a constant to the solution

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

y = (5 t)^{t}

x \to 0 lim (\frac{2 sin ( x ) - sin ( 2 x )}{sin ( x ) - x cos ( x )})

\int_{0}^{4} 3 t^{4} + 3 t + 2 d t

n = 0 \sum \infty (\frac{x}{3})^{n}

\frac{\partial}{\partial x} (e^{3 x yz - 7})

What is the integral of ((x^2-x+24))/(x^3+6x) ?
The integral of ((x^2-x+24))/(x^3+6x) is 4ln|x|-3/2 ln|x^2+6|-1/(sqrt(6))arctan(x/(sqrt(6)))+C