What is the integral from 9 to 16 of 1/(7-sqrt(x)) ?

The integral from 9 to 16 of 1/(7-sqrt(x)) is 2(-1+7(2ln(2)-ln(3)))

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integral from 9 to 16 of 1/(7-sqrt(x))

\int_{9}^{16} \frac{1}{7 - x} d x

2 (- 1 + 7 (2 ln (2) - ln (3)))

Decimal Notation

2.02754 \dots

Solution steps

\int_{9}^{16} \frac{1}{7 - x} d x

Apply u-substitution

: \int_{3}^{4} \frac{2 u}{7 - u} d u

= \int_{3}^{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_{3}^{4} \frac{u}{7 - u} d u

Apply u-substitution

: \int_{4}^{3} - \frac{- v + 7}{v} d v

= 2 \cdot \int_{4}^{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}^{4} - \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}^{4} \frac{- v + 7}{v} d v))

Expand

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

= 2 (- (- \int_{3}^{4} - 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}^{4} 1 d v + \int_{3}^{4} \frac{7}{v} d v)))

\int_{3}^{4} 1 d v = 1

\int_{3}^{4} \frac{7}{v} d v = 7 (2 ln (2) - ln (3))

= 2 (- (- (- 1 + 7 (2 ln (2) - ln (3)))))

Simplify

= 2 (- 1 + 7 (2 ln (2) - ln (3)))

\int \frac{x}{x - 16} d x

n = 0 \sum \infty n^{2} (x)^{n - 1}

z \to - 1 lim (\frac{4 z + 3}{2 z + 1})

5 t^{4}

x^{7} y^{^{'}} = - 8 x^{6} y + cos (2 x)

What is the integral from 9 to 16 of 1/(7-sqrt(x)) ?
The integral from 9 to 16 of 1/(7-sqrt(x)) is 2(-1+7(2ln(2)-ln(3)))