What is the integral from 0 to 9 of 1/(1+sqrt(x)) ?

The integral from 0 to 9 of 1/(1+sqrt(x)) is 2(3-2ln(2))

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integral from 0 to 9 of 1/(1+sqrt(x))

\int_{0}^{9} \frac{1}{1 + x} d x

2 (3 - 2 ln (2))

Decimal

3.22741 \dots

Solution steps

\int_{0}^{9} \frac{1}{1 + x} d x

Apply u-substitution

: \int_{1}^{4} \frac{2 ( u - 1 )}{u} d u

= \int_{1}^{4} \frac{2 ( u - 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}^{4} \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 2 \cdot \int_{1}^{4} 1 - \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

= 2 (\int_{1}^{4} 1 d u - \int_{1}^{4} \frac{1}{u} d u)

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

\int_{1}^{4} \frac{1}{u} d u = 2 ln (2)

= 2 (3 - 2 ln (2))

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What is the integral from 0 to 9 of 1/(1+sqrt(x)) ?
The integral from 0 to 9 of 1/(1+sqrt(x)) is 2(3-2ln(2))