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

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

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

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

- ln ∣ x + 1 ∣ + \frac{6}{x + 1} + 2 ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

= - ln ∣ x + 1 ∣ + \frac{6}{x + 1} + 2 ln ∣ x ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + \frac{6}{x + 1} + 2 ln ∣ x ∣ + C

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