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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{- u + 4}{u} d u

= \int \frac{- u + 4}{u} d u

Expand

\frac{- u + 4}{u} : - 1 + \frac{4}{u}

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

Apply u-substitution

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

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

Take the constant out:

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

= - 4 \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

= - 4 (- \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)

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

Substitute back

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x y - x sin (y))

\frac{( t ^{2} + 1 ) ^{\frac{3}{2}}}{3}^{^{'}}

\frac{d}{d x} ((x + 2) (x - 2)^{3})

- 7 (6 x^{2} + 7)^{- 6}

\frac{c ^{4} - x ^{4}}{c ^{4} + x ^{4}}

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