What is the integral from 3 to 8 of 1/(sqrt(1+x)-1) ?

The integral from 3 to 8 of 1/(sqrt(1+x)-1) is 2(1+ln(2))

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integral from 3 to 8 of 1/(sqrt(1+x)-1)

\int_{3}^{8} \frac{1}{1 + x - 1} d x

2 (1 + ln (2))

Decimal Notation

3.38629 \dots

Solution steps

\int_{3}^{8} \frac{1}{1 + x - 1} d x

Apply u-substitution

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

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

Apply u-substitution

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

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

Expand

\frac{v + 1}{v} : 1 + \frac{1}{v}

= 2 \cdot \int_{1}^{2} 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 (\int_{1}^{2} 1 d v + \int_{1}^{2} \frac{1}{v} d v)

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

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

= 2 (1 + ln (2))

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What is the integral from 3 to 8 of 1/(sqrt(1+x)-1) ?
The integral from 3 to 8 of 1/(sqrt(1+x)-1) is 2(1+ln(2))