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

The integral of (sqrt(4-x))/(sqrt(x)) is 4(arcsin(1/2 sqrt(x))+1/2 sin(2arcsin(1/2 sqrt(x))))+C

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

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

4 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int 4 cos^{2} (v) d v

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

Rewrite using trig identities

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

Take the constant out:

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

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

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 2 \cdot 4 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 2 \cdot 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x)))

Simplify

2 \cdot 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))) : 4 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x)))

= 4 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x)))

Add a constant to the solution

= 4 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))) + C

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