What is the integral from 1 to 4 of 1/(1+2sqrt(x)) ?

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

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integral from 1 to 4 of 1/(1+2sqrt(x))

\int_{1}^{4} \frac{1}{1 + 2 x} d x

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

Decimal

0.74458 \dots

Solution steps

\int_{1}^{4} \frac{1}{1 + 2 x} d x

Apply u-substitution

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

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

Apply u-substitution

: \int_{3}^{5} \frac{v - 1}{4 v} d v

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

Expand

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

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

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

\int_{3}^{5} \frac{1}{v} d v = ln (5) - ln (3)

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

Simplify

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

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

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