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

The integral from 9 to 16 of 1/(9-sqrt(x)) is 2(-1+9(ln(6)-ln(5)))

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

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

2 (- 1 + 9 (ln (6) - ln (5)))

Decimal Notation

1.28178 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{4} \frac{2 u}{9 - u} d u

= \int_{3}^{4} \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_{3}^{4} \frac{u}{9 - u} d u

Apply u-substitution

: \int_{6}^{5} - \frac{- v + 9}{v} d v

= 2 \cdot \int_{6}^{5} - \frac{- v + 9}{v} d v

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

= 2 (- \int_{5}^{6} - \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_{5}^{6} \frac{- v + 9}{v} d v))

Expand

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

= 2 (- (- \int_{5}^{6} - 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_{5}^{6} 1 d v + \int_{5}^{6} \frac{9}{v} d v)))

\int_{5}^{6} 1 d v = 1

\int_{5}^{6} \frac{9}{v} d v = 9 (ln (6) - ln (5))

= 2 (- (- (- 1 + 9 (ln (6) - ln (5)))))

Simplify

= 2 (- 1 + 9 (ln (6) - ln (5)))

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What is the integral from 9 to 16 of 1/(9-sqrt(x)) ?
The integral from 9 to 16 of 1/(9-sqrt(x)) is 2(-1+9(ln(6)-ln(5)))