What is the integral from 4 to 9 of 1/(1-sqrt(x)) ?

The integral from 4 to 9 of 1/(1-sqrt(x)) is 2(-1-ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4 to 9 of 1/(1-sqrt(x))

\int_{4}^{9} \frac{1}{1 - x} d x

2 (- 1 - ln (2))

Decimal

- 3.38629 \dots

Solution steps

\int_{4}^{9} \frac{1}{1 - x} d x

Apply u-substitution

: \int_{- 1}^{- 2} - \frac{2 ( - u + 1 )}{u} d u

= \int_{- 1}^{- 2} - \frac{2 ( - u + 1 )}{u} d u

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

= - \int_{- 2}^{- 1} - \frac{2 ( - u + 1 )}{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}^{- 1} \frac{- u + 1}{u} d u)

Expand

\frac{- u + 1}{u} : - 1 + \frac{1}{u}

= - (- 2 \cdot \int_{- 2}^{- 1} - 1 + \frac{1}{u} 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_{- 2}^{- 1} 1 d u + \int_{- 2}^{- 1} \frac{1}{u} d u))

\int_{- 2}^{- 1} 1 d u = 1

\int_{- 2}^{- 1} \frac{1}{u} d u = - ln (2)

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

Simplify

= 2 (- 1 - ln (2))

\int e^{x^{3} + 1} d x

(1 + x^{2}) y^{^{'}} - 2 x y = 0

x \to 10 lim (\frac{x ^{3}}{3})

(- 3 cos (2 x) - sin (3 x) - 4)^{2}

x \to 2 lim (3 - x + 7)

What is the integral from 4 to 9 of 1/(1-sqrt(x)) ?
The integral from 4 to 9 of 1/(1-sqrt(x)) is 2(-1-ln(2))