What is the integral of 6/(x^3+4x^2+6x) ?

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

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integral of 6/(x^3+4x^2+6x)

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{6 x} d x = \frac{1}{6} ln ∣ x ∣

\int \frac{- x - 4}{6 ( x ^{2} + 4 x + 6 )} d x = \frac{1}{12} (- ln \frac{1}{2} x^{2} + 2 x + 3 - 22 arctan (\frac{1}{2} (x + 2)))

= 6 (\frac{1}{6} ln ∣ x ∣ + \frac{1}{12} (- ln \frac{1}{2} x^{2} + 2 x + 3 - 22 arctan (\frac{1}{2} (x + 2))))

Simplify

6 (\frac{1}{6} ln ∣ x ∣ + \frac{1}{12} (- ln \frac{1}{2} x^{2} + 2 x + 3 - 22 arctan (\frac{1}{2} (x + 2)))) : ln ∣ x ∣ - \frac{1}{2} ln \frac{1}{2} x^{2} + 2 x + 3 - 2 arctan (\frac{1}{2} (x + 2))

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

Add a constant to the solution

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

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