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

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

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

integral

\frac{x ^{4} + x ^{2} + 1}{( x + 1 )}

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

Solution steps

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

Expand

\frac{x ^{4} + x ^{2} + 1}{x + 1} : \frac{x ^{4}}{x + 1} + \frac{x ^{2}}{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{x ^{2}}{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{x ^{2}}{x + 1} d x = \frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + 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 ∣ + \frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣ + ln ∣ x + 1 ∣

Simplify

\frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - x + ln ∣ x + 1 ∣ + \frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣ + ln ∣ x + 1 ∣ : \frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 3 x + 3 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 2

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

Add a constant to the solution

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

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