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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

y = (\frac{x ^{2} + 1}{x ^{2} - 1})^{6}

\frac{d}{d x} (cos (3 x) cos (2 x))

f (x) = 7 - 6 x, at x = - 1

\frac{d}{d x} (ln (tan (\frac{x}{2} + \frac{π}{4})))

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