What is the integral of sqrt((6+x)/(6-x)) ?

The integral of sqrt((6+x)/(6-x)) is -sqrt(x+6)sqrt(6-x)+12arctan(sqrt((x+6)/(6-x)))+C

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integral of sqrt((6+x)/(6-x))

\int \frac{6 + x}{6 - x} d x

- x + 6 6 - x + 12 arctan (\frac{x + 6}{6 - x}) + C

Solution steps

\int \frac{6 + x}{6 - x} d x

Apply u-substitution

= \int - \frac{- u + 12}{u} d u

Take the constant out:

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

= - \int \frac{- u + 12}{u} d u

Expand

\frac{- u + 12}{u} : - 1 + \frac{12}{u}

= - \int - 1 + \frac{12}{u} d u

Apply u-substitution

= - \int - \frac{12 v - 1}{v ^{2}} d v

Take the constant out:

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

= - (- 12 \cdot \int \frac{v - 1}{v ^{2}} d v)

Apply Integration By Parts

= - (- 12 (- \frac{v - 1}{v} - \int - \frac{1}{2 v v - 1} d v))

Take the constant out:

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

= - (- 12 (- \frac{v - 1}{v} - (- \frac{1}{2} \cdot \int \frac{1}{v v - 1} d v)))

Apply u-substitution

= - (- 12 (- \frac{v - 1}{v} - (- \frac{1}{2} \cdot \int \frac{2}{w ^{2} + 1} d w)))

Take the constant out:

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

= - (- 12 (- \frac{v - 1}{v} - (- \frac{1}{2} \cdot 2 \cdot \int \frac{1}{w ^{2} + 1} d w)))

Use the common integral:

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

= - (- 12 (- \frac{v - 1}{v} - (- \frac{1}{2} \cdot 2 arctan (w))))

Substitute back

= - - 12 - \frac{\frac{12}{6 - x} - 1}{\frac{12}{6 - x}} - (- \frac{1}{2} \cdot 2 arctan (\frac{12}{6 - x} - 1))

Simplify

- - 12 - \frac{\frac{12}{6 - x} - 1}{\frac{12}{6 - x}} - (- \frac{1}{2} \cdot 2 arctan (\frac{12}{6 - x} - 1)) : - x + 6 6 - x + 12 arctan (\frac{x + 6}{6 - x})

= - x + 6 6 - x + 12 arctan (\frac{x + 6}{6 - x})

Add a constant to the solution

= - x + 6 6 - x + 12 arctan (\frac{x + 6}{6 - x}) + C

What is the integral of sqrt((6+x)/(6-x)) ?
The integral of sqrt((6+x)/(6-x)) is -sqrt(x+6)sqrt(6-x)+12arctan(sqrt((x+6)/(6-x)))+C