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

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

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

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

\frac{1}{8} ln (3)

Decimal

0.13732 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{( u - 1 ) ^{2}}{8 u} d u

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

Expand

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

= \frac{1}{8} \cdot \int_{1}^{3} 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}^{3} u d u - \int_{1}^{3} 2 d u + \int_{1}^{3} \frac{1}{u} d u)

\int_{1}^{3} u d u = 4

\int_{1}^{3} 2 d u = 4

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

= \frac{1}{8} (4 - 4 + ln (3))

Simplify

\frac{1}{8} (4 - 4 + ln (3)) : \frac{1}{8} ln (3)

= \frac{1}{8} ln (3)

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

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