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

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

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

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

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

Solution steps

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

Expand

\frac{x ^{4} + x + 1}{x ^{2} + x} : \frac{x ^{4}}{x ^{2} + x} + \frac{x}{x ^{2} + x} + \frac{1}{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{x ^{4}}{x ^{2} + x} d x + \int \frac{x}{x ^{2} + x} d x + \int \frac{1}{x ^{2} + x} d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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