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+2ln|x+1|+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 + 2 ln ∣ x + 1 ∣ + 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 + 2 ln ∣ x + 1 ∣

Add a constant to the solution

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

\int \frac{1}{( x - 1 ) ( x - 4 )} d x

\int_{0}^{π} 7 e^{x} sin (x) d x

f (x) = \frac{1}{3 + 2 x}, at x = 2

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

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+2ln|x+1|+C