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

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

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

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

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

Solution steps

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

Expand

\frac{u + 4}{u} : 1 + \frac{4}{u}

= 2 \cdot \int 1 + \frac{4}{u} d u

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 u + \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= 2 (u + 4 ln ∣ u ∣)

Substitute back

u = x - 4

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x^{2} e^{5 x y})

y = - x^{2} - 2 x + 3, y = x^{2} - 1

- 2 (4 e^{- x^{2}} x^{3} - 6 e^{- x^{2}} x)

ln^{2} (2 x)

x \to 1 lim (\frac{∣ x - 1 ∣}{1 - x})

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