What is the integral from 4 to 9 of (sqrt(x))/(sqrt(x)-1) ?

The integral from 4 to 9 of (sqrt(x))/(sqrt(x)-1) is 2ln(2)+7

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integral from 4 to 9 of (sqrt(x))/(sqrt(x)-1)

\int_{4}^{9} \frac{x}{x - 1} d x

2 ln (2) + 7

Decimal

8.38629 \dots

Solution steps

\int_{4}^{9} \frac{x}{x - 1} d x

Apply u-substitution

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

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

Expand

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

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

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

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

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

= 2 (\frac{3}{2} + 2 + ln (2))

Simplify

2 (\frac{3}{2} + 2 + ln (2)) : 2 ln (2) + 7

= 2 ln (2) + 7

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What is the integral from 4 to 9 of (sqrt(x))/(sqrt(x)-1) ?
The integral from 4 to 9 of (sqrt(x))/(sqrt(x)-1) is 2ln(2)+7