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

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

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

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

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

Solution steps

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

Long division

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

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

\int x^{3} d x = \frac{x ^{4}}{4}

\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 + 1} d x = ln ∣ x + 1 ∣

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

Add a constant to the solution

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

\int t t d t

\int (e^{- t}) d t

\int_{\frac{1}{2}}^{1} \frac{y + 4}{y ^{2} + y} d y

y = 0.7 (1.02)^{x}

f (x) = 1 - 4 x^{2}

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