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

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

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

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

2 ln (2) - 1

Decimal

0.38629 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Factor

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

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

Apply u-substitution

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

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

Expand

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

= 2 \cdot \int_{1}^{2} u - 2 + \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

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

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

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

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

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

Simplify

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

= 2 ln (2) - 1

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