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

The integral from 1 to 4 of x/(1+sqrt(x)) is 2(-ln(3)+ln(2)+11/6)

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

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

2 (- ln (3) + ln (2) + \frac{11}{6})

Decimal Notation

2.85573 \dots

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

\int_{2}^{3} v^{2} d v = \frac{19}{3}

\int_{2}^{3} 3 v d v = \frac{15}{2}

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

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

= 2 (\frac{19}{3} - \frac{15}{2} + 3 - (ln (3) - ln (2)))

Simplify

2 (\frac{19}{3} - \frac{15}{2} + 3 - (ln (3) - ln (2))) : 2 (- ln (3) + ln (2) + \frac{11}{6})

= 2 (- ln (3) + ln (2) + \frac{11}{6})

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What is the integral from 1 to 4 of x/(1+sqrt(x)) ?
The integral from 1 to 4 of x/(1+sqrt(x)) is 2(-ln(3)+ln(2)+11/6)