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

The integral from 0 to 1/2 of (x^2)/(2x+1) is (2ln(2)-1)/(16)

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

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

\frac{2 ln ( 2 ) - 1}{16}

Solution steps

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

Apply u-substitution

: \int_{1}^{2} \frac{( u - 1 ) ^{2}}{8 u} d u

= \int_{1}^{2} \frac{( u - 1 ) ^{2}}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{2} \frac{( u - 1 ) ^{2}}{u} d u

Expand

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

= \frac{1}{8} \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

= \frac{1}{8} (\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)

= \frac{1}{8} (\frac{3}{2} - 2 + ln (2))

Simplify

\frac{1}{8} (\frac{3}{2} - 2 + ln (2)) : \frac{2 ln ( 2 ) - 1}{16}

= \frac{2 ln ( 2 ) - 1}{16}

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What is the integral from 0 to 1/2 of (x^2)/(2x+1) ?
The integral from 0 to 1/2 of (x^2)/(2x+1) is (2ln(2)-1)/(16)