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integration of [1/(x-√x)] dx

Hi Aarif,

Here, we can use the substitution;

x = y²

dx/dy = 2y and so dx = 2y dy

Now,

∫[1 / (x-√x)] dx = ∫[1 / (y² - y)] 2y dy

= ∫[1 / y(y - 1)] 2y dy

= ∫[2 / (y - 1)] dy

= 2ln(y - 1) + C

and substituting back for x,

∫[1 / (x-√x)] dx = 2ln(√x - 1) + C

Note that in the real sense the argument of the ln (natural logarithm) function is the absolute value of √x - 1 since the ln of a negative number does not exist.

Regards

Calculus

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