What is the integral of sqrt((4-x)/(2+x)) ?

The integral of sqrt((4-x)/(2+x)) is sqrt(-x+4)sqrt(2+x)-6arctan(sqrt((-x+4)/(2+x)))+C

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integral of sqrt((4-x)/(2+x))

\int \frac{4 - x}{2 + x} d x

- x + 4 2 + x - 6 arctan (\frac{- x + 4}{2 + x}) + C

Solution steps

\int \frac{4 - x}{2 + x} d x

Apply u-substitution

: \int \frac{- u + 6}{u} d u

= \int \frac{- u + 6}{u} d u

Expand

\frac{- u + 6}{u} : - 1 + \frac{6}{u}

= \int - 1 + \frac{6}{u} d u

Apply u-substitution

: \int - \frac{6 v - 1}{v ^{2}} d v

= \int - \frac{6 v - 1}{v ^{2}} d v

Take the constant out:

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

= - 6 \cdot \int \frac{v - 1}{v ^{2}} d v

Apply Integration By Parts

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

= - 6 (- \frac{v - 1}{v} - \int - \frac{1}{2 v v - 1} d v)

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

= - 6 (- \frac{v - 1}{v} - (- arctan (v - 1)))

Substitute back

= - 6 - \frac{\frac{6}{2 + x} - 1}{\frac{6}{2 + x}} - (- arctan (\frac{6}{2 + x} - 1))

Simplify

- 6 - \frac{\frac{6}{2 + x} - 1}{\frac{6}{2 + x}} - (- arctan (\frac{6}{2 + x} - 1)) : - x + 4 2 + x - 6 arctan (\frac{- x + 4}{2 + x})

= - x + 4 2 + x - 6 arctan (\frac{- x + 4}{2 + x})

Add a constant to the solution

= - x + 4 2 + x - 6 arctan (\frac{- x + 4}{2 + x}) + C

\int - 5 x cos (2 x) d x

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\int x - 1 - (x - 1)^{4} d x

What is the integral of sqrt((4-x)/(2+x)) ?
The integral of sqrt((4-x)/(2+x)) is sqrt(-x+4)sqrt(2+x)-6arctan(sqrt((-x+4)/(2+x)))+C