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

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

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

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

x x - 4 + 4 ln (\frac{1}{2} x + x - 4) + C

Solution steps

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

Simplify

\frac{x}{x - 4} : \frac{x}{x - 4}

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

Apply u-substitution

: \int 2 u^{2} + 4 d u

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

Take the constant out:

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

= 2 \cdot \int u^{2} + 4 d u

Apply Trigonometric Substitution

: \int 4 sec^{3} (v) d v

= 2 \cdot \int 4 sec^{3} (v) d v

Take the constant out:

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

= 2 \cdot 4 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 2 \cdot 4 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 \cdot 4 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 2 \cdot 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} x - 4 ) ) sin ( arctan ( \frac{1}{2} x - 4 ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} x - 4)) + sec (arctan (\frac{1}{2} x - 4)))

Simplify

2 \cdot 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} x - 4 ) ) sin ( arctan ( \frac{1}{2} x - 4 ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} x - 4)) + sec (arctan (\frac{1}{2} x - 4))) : x x - 4 + 4 ln (\frac{1}{2} x + x - 4)

= x x - 4 + 4 ln (\frac{1}{2} x + x - 4)

Add a constant to the solution

= x x - 4 + 4 ln (\frac{1}{2} x + x - 4) + C

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