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

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

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

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

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

Solution steps

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

Expand

\frac{4 - x}{x} : \frac{4}{x} - 1

= \int \frac{4}{x} - 1 d x

Apply u-substitution

: \int - \frac{4 u - 1}{u ^{2}} d u

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

Take the constant out:

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

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

Apply Integration By Parts

: - \frac{u - 1}{u} - \int - \frac{1}{2 u u - 1} d u

= - 4 (- \frac{u - 1}{u} - \int - \frac{1}{2 u u - 1} d u)

\int - \frac{1}{2 u u - 1} d u = - arctan (u - 1)

= - 4 (- \frac{u - 1}{u} - (- arctan (u - 1)))

Substitute back

u = \frac{4}{x}

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

Simplify

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

= x 4 - x - 4 arctan (\frac{4 - x}{x})

Add a constant to the solution

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

\frac{d}{d x} (e^{- 6 x} cos (6 x))

(1 + x^{2}) y^{^{'}} + 4 x y = (1 + x^{2})^{- 2}

\frac{d}{d x} (e^{2 x^{2} - 1})

\int_{0}^{8} \frac{1}{64 + x ^{2}} d x

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