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

The integral of (x^2+3x+2)/(x^2(x+3)) is 7/9 ln|x|+2/9 ln|x+3|-2/(3x)+C

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

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

\frac{7}{9} ln ∣ x ∣ + \frac{2}{9} ln ∣ x + 3 ∣ - \frac{2}{3 x} + C

Solution steps

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

Expand

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

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

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

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

\int \frac{2}{x ^{2} ( x + 3 )} d x = 2 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣)

= ln ∣ x + 3 ∣ + ln ∣ x ∣ - ln ∣ x + 3 ∣ + 2 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣)

Simplify

ln ∣ x + 3 ∣ + ln ∣ x ∣ - ln ∣ x + 3 ∣ + 2 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣) : \frac{7}{9} ln ∣ x ∣ + \frac{2}{9} ln ∣ x + 3 ∣ - \frac{2}{3 x}

= \frac{7}{9} ln ∣ x ∣ + \frac{2}{9} ln ∣ x + 3 ∣ - \frac{2}{3 x}

Add a constant to the solution

= \frac{7}{9} ln ∣ x ∣ + \frac{2}{9} ln ∣ x + 3 ∣ - \frac{2}{3 x} + C

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What is the integral of (x^2+3x+2)/(x^2(x+3)) ?
The integral of (x^2+3x+2)/(x^2(x+3)) is 7/9 ln|x|+2/9 ln|x+3|-2/(3x)+C