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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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