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

The integral of 1/(sqrt(2x+1)+\sqrt{2x-1)} is 1/6 ((2x+1)^{3/2}-(2x-1)^{3/2})+C

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

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

\frac{1}{6} ((2 x + 1)^{\frac{3}{2}} - (2 x - 1)^{\frac{3}{2}}) + C

Solution steps

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

Multiply by the conjugate of

2 x + 1 + 2 x - 1 : \frac{2}{2 x + 1 - 2 x - 1}

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

Take the constant out:

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

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

Simplify

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

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

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

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

Simplify

\frac{1}{2} (\frac{1}{3} (2 x + 1)^{\frac{3}{2}} - \frac{1}{3} (2 x - 1)^{\frac{3}{2}}) : \frac{1}{6} ((2 x + 1)^{\frac{3}{2}} - (2 x - 1)^{\frac{3}{2}})

= \frac{1}{6} ((2 x + 1)^{\frac{3}{2}} - (2 x - 1)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{6} ((2 x + 1)^{\frac{3}{2}} - (2 x - 1)^{\frac{3}{2}}) + C

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What is the integral of 1/(sqrt(2x+1)+\sqrt{2x-1)} ?
The integral of 1/(sqrt(2x+1)+\sqrt{2x-1)} is 1/6 ((2x+1)^{3/2}-(2x-1)^{3/2})+C