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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

Add a constant to the solution

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

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