What is the integral from 1 to 9 of 1/(6-sqrt(x)) ?

The integral from 1 to 9 of 1/(6-sqrt(x)) is 2(-2+6(ln(5)-ln(3)))

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integral from 1 to 9 of 1/(6-sqrt(x))

\int_{1}^{9} \frac{1}{6 - x} d x

2 (- 2 + 6 (ln (5) - ln (3)))

Decimal Notation

2.12990 \dots

Solution steps

\int_{1}^{9} \frac{1}{6 - x} d x

Apply u-substitution

: \int_{1}^{3} \frac{2 u}{6 - u} d u

= \int_{1}^{3} \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}^{3} \frac{u}{6 - u} d u

Apply u-substitution

: \int_{5}^{3} - \frac{- v + 6}{v} d v

= 2 \cdot \int_{5}^{3} - \frac{- v + 6}{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 + 6}{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 + 6}{v} d v))

Expand

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

= 2 (- (- \int_{3}^{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_{3}^{5} 1 d v + \int_{3}^{5} \frac{6}{v} d v)))

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

\int_{3}^{5} \frac{6}{v} d v = 6 (ln (5) - ln (3))

= 2 (- (- (- 2 + 6 (ln (5) - ln (3)))))

Simplify

= 2 (- 2 + 6 (ln (5) - ln (3)))

\int \frac{1}{x} 1 + x^{2} d x

f (x) = x^{3} - 2 x + 2, (2, 6)

4 t \frac{d y}{d t} - 3 y = t

\frac{d}{d x} (- 2 x sin (2 x) + cos (2 x))

What is the integral from 1 to 9 of 1/(6-sqrt(x)) ?
The integral from 1 to 9 of 1/(6-sqrt(x)) is 2(-2+6(ln(5)-ln(3)))