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

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

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

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

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

Solution steps

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

Expand

\frac{x ^{4} + x}{x + 1} : \frac{x ^{4}}{x + 1} + \frac{x}{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{x}{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{x}{x + 1} d x = x + 1 - ln ∣ x + 1 ∣

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

Simplify

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

Add a constant to the solution

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

2 x^{3} - x^{2}

e^{3 t} sin (4 t)

\int x^{2} + x d x

n = 0 \sum \infty 9^{n}

f (x) = \frac{x ^{2}}{( x ^{3} ) ( x ^{2} )}

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