What is the integral from 9 to 16 of 1/(2-sqrt(x)) ?

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

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integral from 9 to 16 of 1/(2-sqrt(x))

\int_{9}^{16} \frac{1}{2 - x} d x

2 (- 2 ln (2) - 1)

Decimal Notation

- 4.77258 \dots

Solution steps

\int_{9}^{16} \frac{1}{2 - x} d x

Apply u-substitution

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

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

\frac{u}{2 - u} = \frac{2}{2 - u} - 1

= 2 \cdot \int_{3}^{4} \frac{2}{2 - u} - 1 d u

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} \frac{2}{2 - u} d u - \int_{3}^{4} 1 d u)

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

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

= 2 (- 2 ln (2) - 1)

\frac{\partial}{\partial y} (sin (y) + y, x)

\int_{0}^{2} t e^{t} d t

sin^{2} (x), sin^{3} (x), 0, π

\int 3 sin (x) + 2 - 3 e^{x} d x

f (x) = 4 x (1 - x), g (x) = 4 x - 4

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