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

The integral of 1/(9-sqrt(x)) is -2(-9+sqrt(x)+9ln|9-sqrt(x)|)+C

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integral of 1/(9-sqrt(x))

\int \frac{1}{9 - x} d x

- 2 (- 9 + x + 9 ln 9 - x) + C

Solution steps

\int \frac{1}{9 - x} d x

Apply u-substitution

: \int \frac{2 u}{9 - u} d u

= \int \frac{2 u}{9 - u} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{9 - u} d u

Apply u-substitution

: \int - \frac{- v + 9}{v} d v

= 2 \cdot \int - \frac{- v + 9}{v} d v

Take the constant out:

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

= 2 (- \int \frac{- v + 9}{v} d v)

Expand

\frac{- v + 9}{v} : - 1 + \frac{9}{v}

= 2 (- \int - 1 + \frac{9}{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 1 d v + \int \frac{9}{v} d v))

\int 1 d v = v

\int \frac{9}{v} d v = 9 ln ∣ v ∣

= 2 (- (- v + 9 ln ∣ v ∣))

Substitute back

= 2 (- (- (9 - x) + 9 ln 9 - x))

Simplify

= - 2 (- 9 + x + 9 ln 9 - x)

Add a constant to the solution

= - 2 (- 9 + x + 9 ln 9 - x) + C

\int sin (\frac{3 x}{2}) d x

\int sin^{7} (x) cos^{3} (x) d x

x \to 0 lim (x^{19 x})

\int_{1}^{\infty} f (x) d x

(19 x y^{2} - 15 y) + (19 x^{2} y - 15 x) y^{^{'}} = 0

What is the integral of 1/(9-sqrt(x)) ?
The integral of 1/(9-sqrt(x)) is -2(-9+sqrt(x)+9ln|9-sqrt(x)|)+C