What is the integral from 0 to 2 of (x^3)/(x+1) ?

The integral from 0 to 2 of (x^3)/(x+1) is -ln(3)+8/3

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integral from 0 to 2 of (x^3)/(x+1)

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

- ln (3) + \frac{8}{3}

Decimal Notation

1.56805 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int_{1}^{3} u^{2} d u = \frac{26}{3}

\int_{1}^{3} 3 u d u = 12

\int_{1}^{3} 3 d u = 6

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

= \frac{26}{3} - 12 + 6 - ln (3)

Simplify

\frac{26}{3} - 12 + 6 - ln (3) : - ln (3) + \frac{8}{3}

= - ln (3) + \frac{8}{3}

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What is the integral from 0 to 2 of (x^3)/(x+1) ?
The integral from 0 to 2 of (x^3)/(x+1) is -ln(3)+8/3