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

The integral from 1 to 9 of 1/(9-sqrt(x)) is 2(-2+9ln(4/3))

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

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

2 (- 2 + 9 ln (\frac{4}{3}))

Decimal

1.17827 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{2 u}{9 - u} d u

= \int_{1}^{3} \frac{2 u}{9 - 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}^{3} \frac{u}{9 - u} d u

Apply u-substitution

: \int_{8}^{6} - \frac{- v + 9}{v} d v

= 2 \cdot \int_{8}^{6} - \frac{- v + 9}{v} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{6}^{8} - \frac{- v + 9}{v} d v)

Take the constant out:

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

= 2 (- (- \int_{6}^{8} \frac{- v + 9}{v} d v))

Expand

\frac{- v + 9}{v} : - 1 + \frac{9}{v}

= 2 (- (- \int_{6}^{8} - 1 + \frac{9}{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_{6}^{8} 1 d v + \int_{6}^{8} \frac{9}{v} d v)))

\int_{6}^{8} 1 d v = 2

\int_{6}^{8} \frac{9}{v} d v = 9 ln (\frac{4}{3})

= 2 (- (- (- 2 + 9 ln (\frac{4}{3}))))

Simplify

= 2 (- 2 + 9 ln (\frac{4}{3}))

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What is the integral from 1 to 9 of 1/(9-sqrt(x)) ?
The integral from 1 to 9 of 1/(9-sqrt(x)) is 2(-2+9ln(4/3))