What is the integral from 1 to 16 of 1/(7-sqrt(x)) ?

The integral from 1 to 16 of 1/(7-sqrt(x)) is 2(-3+7ln(2))

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integral from 1 to 16 of 1/(7-sqrt(x))

\int_{1}^{16} \frac{1}{7 - x} d x

2 (- 3 + 7 ln (2))

Solution steps

\int_{1}^{16} \frac{1}{7 - x} d x

Apply u-substitution

: \int_{1}^{4} \frac{2 u}{7 - u} d u

= \int_{1}^{4} \frac{2 u}{7 - 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}{7 - u} d u

Apply u-substitution

: \int_{6}^{3} - \frac{- v + 7}{v} d v

= 2 \cdot \int_{6}^{3} - \frac{- v + 7}{v} d v

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

= 2 (- \int_{3}^{6} - \frac{- v + 7}{v} d v)

Take the constant out:

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

= 2 (- (- \int_{3}^{6} \frac{- v + 7}{v} d v))

Expand

\frac{- v + 7}{v} : - 1 + \frac{7}{v}

= 2 (- (- \int_{3}^{6} - 1 + \frac{7}{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_{3}^{6} 1 d v + \int_{3}^{6} \frac{7}{v} d v)))

\int_{3}^{6} 1 d v = 3

\int_{3}^{6} \frac{7}{v} d v = 7 ln (2)

= 2 (- (- (- 3 + 7 ln (2))))

Simplify

= 2 (- 3 + 7 ln (2))

x \to - 2 + lim (\frac{- 3}{x ^{2} - 4})

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What is the integral from 1 to 16 of 1/(7-sqrt(x)) ?
The integral from 1 to 16 of 1/(7-sqrt(x)) is 2(-3+7ln(2))