What is the integral from 1 to 16 of 1/(8-sqrt(x)) ?

The integral from 1 to 16 of 1/(8-sqrt(x)) is 2(-3+8(ln(7)-2ln(2)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 16 of 1/(8-sqrt(x))

\int_{1}^{16} \frac{1}{8 - x} d x

2 (- 3 + 8 (ln (7) - 2 ln (2)))

Decimal Notation

2.95385 \dots

Solution steps

\int_{1}^{16} \frac{1}{8 - x} d x

Apply u-substitution

: \int_{1}^{4} \frac{2 u}{8 - u} d u

= \int_{1}^{4} \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}^{4} \frac{u}{8 - u} d u

Apply u-substitution

: \int_{7}^{4} - \frac{- v + 8}{v} d v

= 2 \cdot \int_{7}^{4} - \frac{- v + 8}{v} d v

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

= 2 (- \int_{4}^{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_{4}^{7} \frac{- v + 8}{v} d v))

Expand

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

= 2 (- (- \int_{4}^{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_{4}^{7} 1 d v + \int_{4}^{7} \frac{8}{v} d v)))

\int_{4}^{7} 1 d v = 3

\int_{4}^{7} \frac{8}{v} d v = 8 (ln (7) - 2 ln (2))

= 2 (- (- (- 3 + 8 (ln (7) - 2 ln (2)))))

Simplify

= 2 (- 3 + 8 (ln (7) - 2 ln (2)))

\frac{d y}{d x} + 2 x y = y + 4 x - 2

\frac{d}{d x} ((4 x^{3} + 5)^{\frac{3}{2}})

\frac{\partial}{\partial x} (\frac{x ^{2} - 1}{y})

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

\int - 32 x + 12 d x

What is the integral from 1 to 16 of 1/(8-sqrt(x)) ?
The integral from 1 to 16 of 1/(8-sqrt(x)) is 2(-3+8(ln(7)-2ln(2)))