What is the integral of 1/(sqrt(2x)-\sqrt{x+4)} ?

The integral of 1/(sqrt(2x)-\sqrt{x+4)} is

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integral of 1/(sqrt(2x)-\sqrt{x+4)}

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

- 42 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{1}{2} x) - 22 ln \frac{1}{2 2} x + 4 + 1 + 22 ln \frac{1}{2 2} x + 4 - 1 + 2 x + 4 + C

Solution steps

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

Apply radical rule:

assuming

a \geq 0, b \geq 0

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

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

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

Expand

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

Apply the Sum Rule:

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

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

\int \frac{2 x}{( 2 x ) ^{2} - ( x + 4 ) ^{2}} d x = - 42 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{1}{2} x)

\int \frac{x + 4}{( 2 x ) ^{2} - ( x + 4 ) ^{2}} d x = - 22 ln \frac{1}{2 2} x + 4 + 1 + 22 ln \frac{1}{2 2} x + 4 - 1 + 2 x + 4

= - 42 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{1}{2} x) - 22 ln \frac{1}{2 2} x + 4 + 1 + 22 ln \frac{1}{2 2} x + 4 - 1 + 2 x + 4

Add a constant to the solution

= - 42 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{1}{2} x) - 22 ln \frac{1}{2 2} x + 4 + 1 + 22 ln \frac{1}{2 2} x + 4 - 1 + 2 x + 4 + C

What is the integral of 1/(sqrt(2x)-\sqrt{x+4)} ?
The integral of 1/(sqrt(2x)-\sqrt{x+4)} is