What is the integral of (x^5)/(x+1) ?

The integral of (x^5)/(x+1) is (x^5)/5-(x^4)/4+(x^3)/3-(x^2)/2+x-ln|x+1|+C

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integral of (x^5)/(x+1)

\int \frac{x ^{5}}{x + 1} d x

\frac{x ^{5}}{5} - \frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x - ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{x ^{5}}{x + 1} d x

Long division

\frac{x ^{5}}{x + 1} : x^{4} - x^{3} + x^{2} - x + 1 - \frac{1}{x + 1}

= \int x^{4} - 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^{4} 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^{4} d x = \frac{x ^{5}}{5}

\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 ^{5}}{5} - \frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x - ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x - ln ∣ x + 1 ∣ + C

\frac{4 x}{x - 5}

y = tan (\frac{t}{1 + t ^{2}})

\int_{0}^{1} 4 π (x^{2} - x^{10}) d x

\frac{3}{x}

\int 2 x (x^{2} + 5)^{5} d x

What is the integral of (x^5)/(x+1) ?
The integral of (x^5)/(x+1) is (x^5)/5-(x^4)/4+(x^3)/3-(x^2)/2+x-ln|x+1|+C