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

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

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

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

\frac{1}{4} ln ∣ x ∣ + \frac{3}{4} ln ∣ x + 2 ∣ + \frac{x}{x + 2} + \frac{2}{x + 2} - \frac{1}{2 x} + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{x ^{3} + 2 x ^{2}} d x = - \frac{1}{4} ln ∣ x ∣ - \frac{1}{2 x} + \frac{1}{4} ln ∣ x + 2 ∣

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

Simplify

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

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

Add a constant to the solution

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

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