What is the integral from 1 to 9 of 1/(8-sqrt(x)) ?

The integral from 1 to 9 of 1/(8-sqrt(x)) is 2(-2+8(ln(7)-ln(5)))

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integral from 1 to 9 of 1/(8-sqrt(x))

\int_{1}^{9} \frac{1}{8 - x} d x

2 (- 2 + 8 (ln (7) - ln (5)))

Decimal Notation

1.38355 \dots

Solution steps

\int_{1}^{9} \frac{1}{8 - x} d x

Apply u-substitution

: \int_{1}^{3} \frac{2 u}{8 - u} d u

= \int_{1}^{3} \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_{1}^{3} \frac{u}{8 - u} d u

Apply u-substitution

: \int_{7}^{5} - \frac{- v + 8}{v} d v

= 2 \cdot \int_{7}^{5} - \frac{- v + 8}{v} d v

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

= 2 (- \int_{5}^{7} - \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_{5}^{7} \frac{- v + 8}{v} d v))

Expand

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

= 2 (- (- \int_{5}^{7} - 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_{5}^{7} 1 d v + \int_{5}^{7} \frac{8}{v} d v)))

\int_{5}^{7} 1 d v = 2

\int_{5}^{7} \frac{8}{v} d v = 8 (ln (7) - ln (5))

= 2 (- (- (- 2 + 8 (ln (7) - ln (5)))))

Simplify

= 2 (- 2 + 8 (ln (7) - ln (5)))

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f (x) = x cos (x) sin (x)

What is the integral from 1 to 9 of 1/(8-sqrt(x)) ?
The integral from 1 to 9 of 1/(8-sqrt(x)) is 2(-2+8(ln(7)-ln(5)))