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

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

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integral from 0 to 1 of x/(sqrt(x)+1)

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

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

Decimal Notation

0.28037 \dots

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

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

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} v^{2} d v - \int_{1}^{2} 3 v d v + \int_{1}^{2} 3 d v - \int_{1}^{2} \frac{1}{v} d v)

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

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

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

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

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

Combine the fractions

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

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

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