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Calculus/integral calculus

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Question
integration of [1/(x-√x)] dx

Answer
Hi Aarif,
Here, we can use the substitution;
x = y
dx/dy = 2y and so dx = 2y dy
Now,
∫[1 / (x-√x)] dx = ∫[1 / (y - 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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