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

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

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

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

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

Solution steps

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

Expand

\frac{x ^{4} - 1}{x ^{2} + x} : x^{2} - x + 1 - \frac{1}{x}

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

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

Add a constant to the solution

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

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