# How do you evaluate the integral #int 1/(x-1)^(2/3)dx# from 0 to 2?

To evaluate

Here, it is not so at

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The integrand is not defined at

provided that both integrals on the right converge,

provided that both integrals on the right converge.

and

We conclude:

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To evaluate the integral (\int_{0}^{2} \frac{1}{(x-1)^{\frac{2}{3}}} dx), we first need to recognize that this integral involves an improper integral due to the singularity at (x = 1). To handle this, we can make a substitution to transform the integral into a standard form.

Let (u = x - 1), then (du = dx). Also, when (x = 0), (u = -1), and when (x = 2), (u = 1).

Substituting these into the integral, we get:

[ \int_{-1}^{1} \frac{1}{u^{\frac{2}{3}}} du ]

This integral can be evaluated using the power rule for integration.

[ = \left[\frac{3}{1-\frac{2}{3}} u^{1-\frac{2}{3}}\right]_{-1}^{1} ]

[ = \left[ \frac{3}{\frac{1}{3}}u^{\frac{1}{3}} \right]_{-1}^{1} ]

[ = 9\left(\left(1\right)^{\frac{1}{3}} - \left(-1\right)^{\frac{1}{3}}\right) ]

[ = 9\left(1 - (-1)\right) ]

[ = 9(2) ]

[ = 18 ]

Therefore, the value of the integral (\int_{0}^{2} \frac{1}{(x-1)^{\frac{2}{3}}} dx) is (18).

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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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