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

The integral of 1/(4-sqrt(x)) is -2(-4+sqrt(x)+4ln|4-sqrt(x)|)+C

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integral of 1/(4-sqrt(x))

\int \frac{1}{4 - x} d x

- 2 (- 4 + x + 4 ln 4 - x) + C

Solution steps

\int \frac{1}{4 - x} d x

Apply u-substitution

: \int \frac{2 u}{4 - u} d u

= \int \frac{2 u}{4 - 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}{4 - u} d u

Apply u-substitution

: \int - \frac{- v + 4}{v} d v

= 2 \cdot \int - \frac{- v + 4}{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 + 4}{v} d v)

Expand

\frac{- v + 4}{v} : - 1 + \frac{4}{v}

= 2 (- \int - 1 + \frac{4}{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{4}{v} d v))

\int 1 d v = v

\int \frac{4}{v} d v = 4 ln ∣ v ∣

= 2 (- (- v + 4 ln ∣ v ∣))

Substitute back

= 2 (- (- (4 - x) + 4 ln 4 - x))

Simplify

= - 2 (- 4 + x + 4 ln 4 - x)

Add a constant to the solution

= - 2 (- 4 + x + 4 ln 4 - x) + C

x \to 1 lim (x^{\frac{6}{1 - x}})

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

x \to 1 lim (4 x^{3})

x \to - 6 lim (5)

What is the integral of 1/(4-sqrt(x)) ?
The integral of 1/(4-sqrt(x)) is -2(-4+sqrt(x)+4ln|4-sqrt(x)|)+C