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

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

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

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

\frac{1}{3} ln x^{2} + 6 + \frac{1}{3} ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

Simplify

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

= \frac{1}{3} ln x^{2} + 6 + \frac{1}{3} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{3} ln x^{2} + 6 + \frac{1}{3} ln ∣ x ∣ + C

y = x^{2} + 2

\int \frac{2}{x ^{2} + 16} d x

\frac{d}{d x} (\frac{x ^{6} + 2}{8 x ^{2}})

\frac{5 u ^{2}}{( u ^{2} + u ) ^{3}}

\frac{\partial}{\partial x} (a cos (k x))

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