What is the integral of (4x^2+3x+2)/(x^2(x+1)) ?

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

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integral of (4x^2+3x+2)/(x^2(x+1))

\int \frac{4 x ^{2} + 3 x + 2}{x ^{2} ( x + 1 )} d x

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

Solution steps

\int \frac{4 x ^{2} + 3 x + 2}{x ^{2} ( x + 1 )} d x

Expand

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

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

\int \frac{4 x ^{2}}{x ^{2} ( x + 1 )} d x = 4 ln ∣ x + 1 ∣

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

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

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

Simplify

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

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

Add a constant to the solution

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

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