What is the integral from 4 to 16 of 1/(7-sqrt(x)) ?

The integral from 4 to 16 of 1/(7-sqrt(x)) is 2(-2+7(ln(5)-ln(3)))

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integral from 4 to 16 of 1/(7-sqrt(x))

\int_{4}^{16} \frac{1}{7 - x} d x

2 (- 2 + 7 (ln (5) - ln (3)))

Decimal Notation

3.15155 \dots

Solution steps

\int_{4}^{16} \frac{1}{7 - x} d x

Apply u-substitution

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

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

Apply u-substitution

: \int_{5}^{3} - \frac{- v + 7}{v} d v

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

Expand

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

= 2 (- (- \int_{3}^{5} - 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}^{5} 1 d v + \int_{3}^{5} \frac{7}{v} d v)))

\int_{3}^{5} 1 d v = 2

\int_{3}^{5} \frac{7}{v} d v = 7 (ln (5) - ln (3))

= 2 (- (- (- 2 + 7 (ln (5) - ln (3)))))

Simplify

= 2 (- 2 + 7 (ln (5) - ln (3)))

\frac{\partial}{\partial x} (z + yz sin (2 x yz))

\int_{- 4}^{- 2} (- x + ∣ - 3 x - 9 ∣) d x

\frac{d y}{d x} = 2 y - y^{2}

\frac{d}{d x} (2 x^{7} - 3 x^{6} + 3 x^{3} - 4 x^{2} - 7)

\int \frac{7 - 7 x}{1 - x} d x

What is the integral from 4 to 16 of 1/(7-sqrt(x)) ?
The integral from 4 to 16 of 1/(7-sqrt(x)) is 2(-2+7(ln(5)-ln(3)))