What is the integral from 4 to 16 of 1/(8-sqrt(x)) ?

The integral from 4 to 16 of 1/(8-sqrt(x)) is 2(-2+8ln(3/2))

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integral from 4 to 16 of 1/(8-sqrt(x))

\int_{4}^{16} \frac{1}{8 - x} d x

2 (- 2 + 8 ln (\frac{3}{2}))

Decimal Notation

2.48744 \dots

Solution steps

\int_{4}^{16} \frac{1}{8 - x} d x

Apply u-substitution

: \int_{2}^{4} \frac{2 u}{8 - u} d u

= \int_{2}^{4} \frac{2 u}{8 - 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}^{4} \frac{u}{8 - u} d u

Apply u-substitution

: \int_{6}^{4} - \frac{- v + 8}{v} d v

= 2 \cdot \int_{6}^{4} - \frac{- v + 8}{v} d v

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

= 2 (- \int_{4}^{6} - \frac{- v + 8}{v} d v)

Take the constant out:

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

= 2 (- (- \int_{4}^{6} \frac{- v + 8}{v} d v))

Expand

\frac{- v + 8}{v} : - 1 + \frac{8}{v}

= 2 (- (- \int_{4}^{6} - 1 + \frac{8}{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_{4}^{6} 1 d v + \int_{4}^{6} \frac{8}{v} d v)))

\int_{4}^{6} 1 d v = 2

\int_{4}^{6} \frac{8}{v} d v = 8 ln (\frac{3}{2})

= 2 (- (- (- 2 + 8 ln (\frac{3}{2}))))

Simplify

= 2 (- 2 + 8 ln (\frac{3}{2}))

x \to 2 + lim (\frac{8}{x - 2})

f (x) = 8 x^{2} - x^{3}, (2, 24)

(3, 4), (5, 8)

y^{^{'}} = \frac{2 y}{3 x}

\int \frac{x + 6}{x ^{2} + 11 x + 18} d x

What is the integral from 4 to 16 of 1/(8-sqrt(x)) ?
The integral from 4 to 16 of 1/(8-sqrt(x)) is 2(-2+8ln(3/2))