What is the integral of sqrt((3-x)/(3+x)) ?

The integral of sqrt((3-x)/(3+x)) is -6(-(sqrt(6/(3+x)-1))/(6/(3+x))-(-arctan(sqrt(6/(3+x)-1))))+C

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integral of sqrt((3-x)/(3+x))

\int \frac{3 - x}{3 + x} d x

- 6 - \frac{\frac{6}{3 + x} - 1}{\frac{6}{3 + x}} - (- arctan (\frac{6}{3 + x} - 1)) + C

Solution steps

\int \frac{3 - x}{3 + 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}{3 + x} - 1}{\frac{6}{3 + x}} - (- arctan (\frac{6}{3 + x} - 1))

Add a constant to the solution

= - 6 - \frac{\frac{6}{3 + x} - 1}{\frac{6}{3 + x}} - (- arctan (\frac{6}{3 + x} - 1)) + C

\int \frac{ln ∣ x ∣}{x} d x

\int \frac{3}{x 9 - ( ln ( x ) ) ^{2}} d x

f (x) = x^{2} - 10 x + 2, at x = 0

x \to 0 lim (csc (x) - cot (x))

\int 0.1 e^{- 0.1 x} d x

What is the integral of sqrt((3-x)/(3+x)) ?
The integral of sqrt((3-x)/(3+x)) is -6(-(sqrt(6/(3+x)-1))/(6/(3+x))-(-arctan(sqrt(6/(3+x)-1))))+C