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

The integral of (sqrt(x-3))/(x+3) is 2sqrt(6)(-arctan(sqrt(1/6 (x-3)))+sqrt(1/6 (x-3)))+C

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} + 6} d v

= \int \frac{2 v ^{2}}{v ^{2} + 6} d v

Take the constant out:

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

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

\frac{v ^{2}}{v ^{2} + 6} = - \frac{6}{v ^{2} + 6} + 1

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

Apply Integral Substitution

= 2 \cdot \int 6 (- \frac{1}{w ^{2} + 1} + 1) d w

Take the constant out:

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

= 26 \cdot \int - \frac{1}{w ^{2} + 1} + 1 d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 26 (- \int \frac{1}{w ^{2} + 1} d w + \int 1 d w)

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

\int 1 d w = w

= 26 (- arctan (w) + w)

Substitute back

= 26 (- arctan (\frac{x + 3 - 6}{6}) + \frac{x + 3 - 6}{6})

Simplify

26 (- arctan (\frac{x + 3 - 6}{6}) + \frac{x + 3 - 6}{6}) : 26 (- arctan (\frac{1}{6} (x - 3)) + \frac{1}{6} (x - 3))

= 26 (- arctan (\frac{1}{6} (x - 3)) + \frac{1}{6} (x - 3))

Add a constant to the solution

= 26 (- arctan (\frac{1}{6} (x - 3)) + \frac{1}{6} (x - 3)) + C

\frac{d}{d x} (\frac{1}{2} x^{4} - 4 x^{2} + 4)

y = 6 x

\frac{\partial}{\partial z} (x z - 7 yz + 4)

\int e^{- s t} cos (2 t) d t

x^{2} + 3

What is the integral of (sqrt(x-3))/(x+3) ?
The integral of (sqrt(x-3))/(x+3) is 2sqrt(6)(-arctan(sqrt(1/6 (x-3)))+sqrt(1/6 (x-3)))+C