What is the integral from 1 to 4 of x/(3+sqrt(x)) ?

The integral from 1 to 4 of x/(3+sqrt(x)) is 2(-27(ln(5)-2ln(2))+41/6)

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integral from 1 to 4 of x/(3+sqrt(x))

\int_{1}^{4} \frac{x}{3 + x} d x

2 (- 27 (ln (5) - 2 ln (2)) + \frac{41}{6})

Decimal Notation

1.61691 \dots

Solution steps

\int_{1}^{4} \frac{x}{3 + x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Long division

\frac{u ^{3}}{3 + u} : u^{2} - 3 u + 9 - \frac{27}{u + 3}

= 2 \cdot \int_{1}^{2} u^{2} - 3 u + 9 - \frac{27}{u + 3} 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^{2} d u - \int_{1}^{2} 3 u d u + \int_{1}^{2} 9 d u - \int_{1}^{2} \frac{27}{u + 3} d u)

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

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

\int_{1}^{2} 9 d u = 9

\int_{1}^{2} \frac{27}{u + 3} d u = 27 (ln (5) - 2 ln (2))

= 2 (\frac{7}{3} - \frac{9}{2} + 9 - 27 (ln (5) - 2 ln (2)))

Combine the fractions

9 + \frac{7}{3} - \frac{9}{2} : \frac{41}{6}

= 2 (- 27 (ln (5) - 2 ln (2)) + \frac{41}{6})

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What is the integral from 1 to 4 of x/(3+sqrt(x)) ?
The integral from 1 to 4 of x/(3+sqrt(x)) is 2(-27(ln(5)-2ln(2))+41/6)