What is the integral of (sqrt(x))/(sqrt(x)-2) ?

The integral of (sqrt(x))/(sqrt(x)-2) is x+4sqrt(x)+8ln|sqrt(x)-2|-12+C

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integral of (sqrt(x))/(sqrt(x)-2)

\int \frac{x}{x - 2} d x

x + 4 x + 8 ln x - 2 - 12 + C

Solution steps

\int \frac{x}{x - 2} d x

Apply u-substitution

: \int \frac{2 ( u ^{2} + 4 u + 4 )}{u} d u

= \int \frac{2 ( u ^{2} + 4 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 ^{2} + 4 u + 4}{u} d u

Expand

\frac{u ^{2} + 4 u + 4}{u} : u + 4 + \frac{4}{u}

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

\int u d u = \frac{u ^{2}}{2}

\int 4 d u = 4 u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= 2 (\frac{u ^{2}}{2} + 4 u + 4 ln ∣ u ∣)

Substitute back

u = x - 2

= 2 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln x - 2)

Simplify

2 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln x - 2) : x + 4 x + 8 ln x - 2 - 12

= x + 4 x + 8 ln x - 2 - 12

Add a constant to the solution

= x + 4 x + 8 ln x - 2 - 12 + C

x \to - 2 lim (\frac{5 x ^{2} + 6 x - 8}{x + 2})

\frac{d}{d x} ((x^{2} - 9)^{2})

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

\frac{8}{s ^{4}}

x y + y^{^{'}} = 2 x

What is the integral of (sqrt(x))/(sqrt(x)-2) ?
The integral of (sqrt(x))/(sqrt(x)-2) is x+4sqrt(x)+8ln|sqrt(x)-2|-12+C