# Evaluate the integral #int \ sqrt(x-x^2)/x \ dx #?

This can be easily verified by differentiating the LHS.

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# int \ sqrt(x-x^2)/x \ dx = 1/2 arcsin(2x-1)+sqrt(x - x^2) + C#

We seek:

We can start by completing the square of the numerator to get:

So we can write:

Substituting this into the integral and we get:

And from the substitution we have:

And so we can reverse the substitution to get:

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The integral of (\int \frac{\sqrt{x-x^2}}{x} , dx) can be evaluated as follows:

[ \int \frac{\sqrt{x-x^2}}{x} , dx = \frac{\pi}{2} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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