What is the integral from 0 to 1 of (x^3+1)/(x+1) ?

The integral from 0 to 1 of (x^3+1)/(x+1) is 5/6

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integral from 0 to 1 of (x^3+1)/(x+1)

\int_{0}^{1} \frac{x ^{3} + 1}{x + 1} d x

\frac{5}{6}

Decimal Notation

0.83333 \dots

Solution steps

\int_{0}^{1} \frac{x ^{3} + 1}{x + 1} d x

Simplify

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

= \int_{0}^{1} x^{2} - 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_{0}^{1} x^{2} d x - \int_{0}^{1} x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 1 d x = 1

= \frac{1}{3} - \frac{1}{2} + 1

Simplify

\frac{1}{3} - \frac{1}{2} + 1 : \frac{5}{6}

= \frac{5}{6}

ln (\frac{1}{1 + e ^{- x}})

\frac{d}{d x} (3 x e^{- x} - 2 x)

\frac{d}{d x} (\frac{x + 1 + x}{x ^{2} - 1})

\frac{d}{d x} (2 x + \frac{18}{x} + 1)

\int (2 x^{\frac{1}{2}}) d x

What is the integral from 0 to 1 of (x^3+1)/(x+1) ?
The integral from 0 to 1 of (x^3+1)/(x+1) is 5/6