What is the integral from 9 to 16 of 1/(sqrt(x)-1) ?

The integral from 9 to 16 of 1/(sqrt(x)-1) is 2(1+ln(3)-ln(2))

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integral from 9 to 16 of 1/(sqrt(x)-1)

\int_{9}^{16} \frac{1}{x - 1} d x

2 (1 + ln (3) - ln (2))

Decimal

2.81093 \dots

Solution steps

\int_{9}^{16} \frac{1}{x - 1} d x

Apply u-substitution

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

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

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

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

\int_{2}^{3} 1 d u = 1

\int_{2}^{3} \frac{1}{u} d u = ln (3) - ln (2)

= 2 (1 + ln (3) - ln (2))

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What is the integral from 9 to 16 of 1/(sqrt(x)-1) ?
The integral from 9 to 16 of 1/(sqrt(x)-1) is 2(1+ln(3)-ln(2))