What is the integral from 1 to 4 of 1/(6-sqrt(x)) ?

The integral from 1 to 4 of 1/(6-sqrt(x)) is 2(-1+6(ln(5)-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 4 of 1/(6-sqrt(x))

\int_{1}^{4} \frac{1}{6 - x} d x

2 (- 1 + 6 (ln (5) - 2 ln (2)))

Decimal Notation

0.67772 \dots

Solution steps

\int_{1}^{4} \frac{1}{6 - x} d x

Apply u-substitution

: \int_{1}^{2} \frac{2 u}{6 - u} d u

= \int_{1}^{2} \frac{2 u}{6 - 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}^{2} \frac{u}{6 - u} d u

Apply u-substitution

: \int_{5}^{4} - \frac{- v + 6}{v} d v

= 2 \cdot \int_{5}^{4} - \frac{- v + 6}{v} d v

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

= 2 (- \int_{4}^{5} - \frac{- v + 6}{v} d v)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 (- (- \int_{4}^{5} \frac{- v + 6}{v} d v))

Expand

\frac{- v + 6}{v} : - 1 + \frac{6}{v}

= 2 (- (- \int_{4}^{5} - 1 + \frac{6}{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}^{5} 1 d v + \int_{4}^{5} \frac{6}{v} d v)))

\int_{4}^{5} 1 d v = 1

\int_{4}^{5} \frac{6}{v} d v = 6 (ln (5) - 2 ln (2))

= 2 (- (- (- 1 + 6 (ln (5) - 2 ln (2)))))

Simplify

= 2 (- 1 + 6 (ln (5) - 2 ln (2)))

f (x) = arcsin (x^{2})

n = 1 \sum \infty \frac{n}{2 n}

f (x) = (x^{2} - 8) (x^{2} + 9)

\int x - 3 d x

\int \frac{( sin ( x ) )}{x} d x

What is the integral from 1 to 4 of 1/(6-sqrt(x)) ?
The integral from 1 to 4 of 1/(6-sqrt(x)) is 2(-1+6(ln(5)-2ln(2)))