What is the integral from 0 to 4 of x/(1+sqrt(x)) ?

The integral from 0 to 4 of x/(1+sqrt(x)) is 2(-ln(3)+8/3)

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integral from 0 to 4 of x/(1+sqrt(x))

\int_{0}^{4} \frac{x}{1 + x} d x

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

Decimal

3.13610 \dots

Solution steps

\int_{0}^{4} \frac{x}{1 + x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

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

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

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

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

= 2 (\frac{26}{3} - 12 + 6 - ln (3))

Simplify

2 (\frac{26}{3} - 12 + 6 - ln (3)) : 2 (- ln (3) + \frac{8}{3})

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

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What is the integral from 0 to 4 of x/(1+sqrt(x)) ?
The integral from 0 to 4 of x/(1+sqrt(x)) is 2(-ln(3)+8/3)