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

The integral of sqrt((4-x)/(4+x)) is -8(-(sqrt(8/(4+x)-1))/(8/(4+x))-(-arctan(sqrt(8/(4+x)-1))))+C

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

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

- 8 - \frac{\frac{8}{4 + x} - 1}{\frac{8}{4 + x}} - (- arctan (\frac{8}{4 + x} - 1)) + C

Solution steps

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

Apply u-substitution

: \int \frac{- u + 8}{u} d u

= \int \frac{- u + 8}{u} d u

Expand

\frac{- u + 8}{u} : - 1 + \frac{8}{u}

= \int - 1 + \frac{8}{u} d u

Apply u-substitution

: \int - \frac{8 v - 1}{v ^{2}} d v

= \int - \frac{8 v - 1}{v ^{2}} d v

Take the constant out:

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

= - 8 \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

= - 8 (- \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)

= - 8 (- \frac{v - 1}{v} - (- arctan (v - 1)))

Substitute back

= - 8 - \frac{\frac{8}{4 + x} - 1}{\frac{8}{4 + x}} - (- arctan (\frac{8}{4 + x} - 1))

Add a constant to the solution

= - 8 - \frac{\frac{8}{4 + x} - 1}{\frac{8}{4 + x}} - (- arctan (\frac{8}{4 + x} - 1)) + C

\int (y^{\frac{2}{3}} - 4 y^{- \frac{1}{5}} + 4) d y

\frac{d}{d x} (ln (\frac{x + 1}{x - 3}))

f (x) = \frac{5}{x ^{3}}

\int 8 u^{\frac{7}{2}} d u

5^{x}, 0

What is the integral of sqrt((4-x)/(4+x)) ?
The integral of sqrt((4-x)/(4+x)) is -8(-(sqrt(8/(4+x)-1))/(8/(4+x))-(-arctan(sqrt(8/(4+x)-1))))+C