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

The integral from 0 to 1 of 1/(3+2sqrt(x)) is (-3ln(5)+3ln(3))/2+1

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integral from 0 to 1 of 1/(3+2sqrt(x))

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

\frac{- 3 ln ( 5 ) + 3 ln ( 3 )}{2} + 1

Decimal

0.23376 \dots

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int_{3}^{5} \frac{v - 3}{4 v} d v

= 2 \cdot \int_{3}^{5} \frac{v - 3}{4 v} d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int_{3}^{5} \frac{v - 3}{v} d v

Expand

\frac{v - 3}{v} : 1 - \frac{3}{v}

= 2 \cdot \frac{1}{4} \cdot \int_{3}^{5} 1 - \frac{3}{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 \cdot \frac{1}{4} (\int_{3}^{5} 1 d v - \int_{3}^{5} \frac{3}{v} d v)

\int_{3}^{5} 1 d v = 2

\int_{3}^{5} \frac{3}{v} d v = 3 (ln (5) - ln (3))

= 2 \cdot \frac{1}{4} (2 - 3 (ln (5) - ln (3)))

Simplify

2 \cdot \frac{1}{4} (2 - 3 (ln (5) - ln (3))) : \frac{- 3 ln ( 5 ) + 3 ln ( 3 )}{2} + 1

= \frac{- 3 ln ( 5 ) + 3 ln ( 3 )}{2} + 1

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What is the integral from 0 to 1 of 1/(3+2sqrt(x)) ?
The integral from 0 to 1 of 1/(3+2sqrt(x)) is (-3ln(5)+3ln(3))/2+1