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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

\frac{u ^{2}}{u ^{2} + 4} = - \frac{4}{u ^{2} + 4} + 1

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

Apply Integral Substitution

= 2 \cdot \int 2 (- \frac{1}{v ^{2} + 1} + 1) d v

Take the constant out:

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

= 2 \cdot 2 \cdot \int - \frac{1}{v ^{2} + 1} + 1 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 2 (- \int \frac{1}{v ^{2} + 1} d v + \int 1 d v)

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= 2 \cdot 2 (- arctan (v) + v)

Substitute back

= 2 \cdot 2 (- arctan (\frac{x - 4}{2}) + \frac{x - 4}{2})

Simplify

2 \cdot 2 (- arctan (\frac{x - 4}{2}) + \frac{x - 4}{2}) : 4 (- arctan (\frac{1}{2} x - 4) + \frac{1}{2} x - 4)

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

Add a constant to the solution

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

y^{^{'}} = \frac{7 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2}}{x ^{5} y}

(- 8, - 3), (- 3, 4)

f (x) = \frac{5}{x}, at x = 1

\frac{1}{2 s + 3}

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

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