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

The integral from 1 to 16 of 1/(9-sqrt(x)) is 2(-3+9(3ln(2)-ln(5)))

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

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

2 (- 3 + 9 (3 ln (2) - ln (5)))

Decimal Notation

2.46006 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{4} \frac{2 u}{9 - u} d u

= \int_{1}^{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_{1}^{4} \frac{u}{9 - u} d u

Apply u-substitution

: \int_{8}^{5} - \frac{- v + 9}{v} d v

= 2 \cdot \int_{8}^{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}^{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_{5}^{8} \frac{- v + 9}{v} d v))

Expand

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

= 2 (- (- \int_{5}^{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_{5}^{8} 1 d v + \int_{5}^{8} \frac{9}{v} d v)))

\int_{5}^{8} 1 d v = 3

\int_{5}^{8} \frac{9}{v} d v = 9 (3 ln (2) - ln (5))

= 2 (- (- (- 3 + 9 (3 ln (2) - ln (5)))))

Simplify

= 2 (- 3 + 9 (3 ln (2) - ln (5)))

f (x) = tan (3 x), at x = \frac{π}{9}

\int 2 + 2 cos (4 x) d x

\frac{\partial}{\partial y} (\frac{x}{( x ^{2} + y ^{2} )})

\frac{d}{d x} (sin (\frac{2 x}{x ^{4} - 4 x}))

n = 0 \sum \infty \frac{9}{n ( n + 3 )}

What is the integral from 1 to 16 of 1/(9-sqrt(x)) ?
The integral from 1 to 16 of 1/(9-sqrt(x)) is 2(-3+9(3ln(2)-ln(5)))