What is the integral from 9 to 16 of 1/(6-sqrt(x)) ?

The integral from 9 to 16 of 1/(6-sqrt(x)) is 2(-1+6(ln(3)-ln(2)))

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integral from 9 to 16 of 1/(6-sqrt(x))

\int_{9}^{16} \frac{1}{6 - x} d x

2 (- 1 + 6 (ln (3) - ln (2)))

Decimal

2.86558 \dots

Solution steps

\int_{9}^{16} \frac{1}{6 - x} d x

Apply u-substitution

: \int_{3}^{4} \frac{2 u}{6 - u} d u

= \int_{3}^{4} \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_{3}^{4} \frac{u}{6 - u} d u

Apply u-substitution

: \int_{3}^{2} - \frac{- v + 6}{v} d v

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

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

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

Expand

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

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

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

\int_{2}^{3} \frac{6}{v} d v = 6 (ln (3) - ln (2))

= 2 (- (- (- 1 + 6 (ln (3) - ln (2)))))

Simplify

= 2 (- 1 + 6 (ln (3) - ln (2)))

\frac{d}{d x} (\frac{ln ( x ) - 10}{ln ( 10 )})

\int_{2}^{e^{4}} \frac{1}{x} d x

\frac{4}{s ^{4}}

\int x^{7} e^{3 x^{4}} d x

\frac{t - t}{t ^{\frac{1}{9}}}

What is the integral from 9 to 16 of 1/(6-sqrt(x)) ?
The integral from 9 to 16 of 1/(6-sqrt(x)) is 2(-1+6(ln(3)-ln(2)))