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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{3 x} d x = \frac{2}{3} ln ∣ x ∣

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

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

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

Add a constant to the solution

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

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