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

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

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

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

2 ln (2)

Decimal

1.38629 \dots

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int_{2}^{4} \frac{1}{v} d v

= 2 \cdot \int_{2}^{4} \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 [ln ∣ v ∣]_{2}^{4}

Compute the boundaries:

ln (2)

= 2 ln (2)

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