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

The integral of (sqrt(x))/(sqrt(x)-4) is x+8sqrt(x)+32ln|sqrt(x)-4|-48+C

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

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

x + 8 x + 32 ln x - 4 - 48 + C

Solution steps

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

Apply u-substitution

: \int \frac{2 ( u ^{2} + 8 u + 16 )}{u} d u

= \int \frac{2 ( u ^{2} + 8 u + 16 )}{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} + 8 u + 16}{u} d u

Expand

\frac{u ^{2} + 8 u + 16}{u} : u + 8 + \frac{16}{u}

= 2 \cdot \int u + 8 + \frac{16}{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 8 d u + \int \frac{16}{u} d u)

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

\int 8 d u = 8 u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= 2 (\frac{u ^{2}}{2} + 8 u + 16 ln ∣ u ∣)

Substitute back

u = x - 4

= 2 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln x - 4)

Simplify

2 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln x - 4) : x + 8 x + 32 ln x - 4 - 48

= x + 8 x + 32 ln x - 4 - 48

Add a constant to the solution

= x + 8 x + 32 ln x - 4 - 48 + C

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What is the integral of (sqrt(x))/(sqrt(x)-4) ?
The integral of (sqrt(x))/(sqrt(x)-4) is x+8sqrt(x)+32ln|sqrt(x)-4|-48+C