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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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