What is the integral of 1/(x-2sqrt(x)) ?

The integral of 1/(x-2sqrt(x)) is ln|x-4|-ln| 1/2 sqrt(x)+1|+ln| 1/2 sqrt(x)-1|+C

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integral of 1/(x-2sqrt(x))

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

ln ∣ x - 4 ∣ - ln \frac{1}{2} x + 1 + ln \frac{1}{2} x - 1 + C

Solution steps

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

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

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

Expand

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

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

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

= ln ∣ x - 4 ∣ - ln \frac{1}{2} x + 1 + ln \frac{1}{2} x - 1

Add a constant to the solution

= ln ∣ x - 4 ∣ - ln \frac{1}{2} x + 1 + ln \frac{1}{2} x - 1 + C

What is the integral of 1/(x-2sqrt(x)) ?
The integral of 1/(x-2sqrt(x)) is ln|x-4|-ln| 1/2 sqrt(x)+1|+ln| 1/2 sqrt(x)-1|+C