What is the integral of 1/(sqrt(3x+2)-\sqrt{3x-2)} ?

The integral of 1/(sqrt(3x+2)-\sqrt{3x-2)} is 1/18 ((3x+2)^{3/2}+(3x-2)^{3/2})+C

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integral of 1/(sqrt(3x+2)-\sqrt{3x-2)}

\int \frac{1}{3 x + 2 - 3 x - 2} d x

\frac{1}{18} ((3 x + 2)^{\frac{3}{2}} + (3 x - 2)^{\frac{3}{2}}) + C

Solution steps

\int \frac{1}{3 x + 2 - 3 x - 2} d x

Multiply by the conjugate of

3 x + 2 - 3 x - 2 : \frac{4}{3 x + 2 + 3 x - 2}

= \int \frac{1}{\frac{4}{3 x + 2 + 3 x - 2}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{\frac{1}{3 x + 2 + 3 x - 2}} d x

Simplify

= \frac{1}{4} \cdot \int 3 x + 2 + 3 x - 2 d x

Apply the Sum Rule:

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

= \frac{1}{4} (\int 3 x + 2 d x + \int 3 x - 2 d x)

\int 3 x + 2 d x = \frac{2}{9} (3 x + 2)^{\frac{3}{2}}

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

= \frac{1}{4} (\frac{2}{9} (3 x + 2)^{\frac{3}{2}} + \frac{2}{9} (3 x - 2)^{\frac{3}{2}})

Simplify

\frac{1}{4} (\frac{2}{9} (3 x + 2)^{\frac{3}{2}} + \frac{2}{9} (3 x - 2)^{\frac{3}{2}}) : \frac{1}{18} ((3 x + 2)^{\frac{3}{2}} + (3 x - 2)^{\frac{3}{2}})

= \frac{1}{18} ((3 x + 2)^{\frac{3}{2}} + (3 x - 2)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{18} ((3 x + 2)^{\frac{3}{2}} + (3 x - 2)^{\frac{3}{2}}) + C

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What is the integral of 1/(sqrt(3x+2)-\sqrt{3x-2)} ?
The integral of 1/(sqrt(3x+2)-\sqrt{3x-2)} is 1/18 ((3x+2)^{3/2}+(3x-2)^{3/2})+C