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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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