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

The integral from 0 to 1 of (x^3)/(1+x) is -ln(2)+5/6

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

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

- ln (2) + \frac{5}{6}

Decimal Notation

0.14018 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} \frac{( u - 1 ) ^{3}}{u} d u

= \int_{1}^{2} \frac{( u - 1 ) ^{3}}{u} d u

Expand

\frac{( u - 1 ) ^{3}}{u} : u^{2} - 3 u + 3 - \frac{1}{u}

= \int_{1}^{2} u^{2} - 3 u + 3 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} u^{2} d u - \int_{1}^{2} 3 u d u + \int_{1}^{2} 3 d u - \int_{1}^{2} \frac{1}{u} d u

\int_{1}^{2} u^{2} d u = \frac{7}{3}

\int_{1}^{2} 3 u d u = \frac{9}{2}

\int_{1}^{2} 3 d u = 3

\int_{1}^{2} \frac{1}{u} d u = ln (2)

= \frac{7}{3} - \frac{9}{2} + 3 - ln (2)

Combine the fractions

3 + \frac{7}{3} - \frac{9}{2} : \frac{5}{6}

= - ln (2) + \frac{5}{6}

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What is the integral from 0 to 1 of (x^3)/(1+x) ?
The integral from 0 to 1 of (x^3)/(1+x) is -ln(2)+5/6