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

The integral is not convergent

The integral is not convergent.

Now we have that:

and

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To evaluate the integral ( \int_{0}^{1} \frac{1}{1-x} , dx ), we can use a basic antiderivative and the Fundamental Theorem of Calculus.

First, we find the antiderivative of ( \frac{1}{1-x} ), which is ( -\ln|1-x| ).

Next, we evaluate this antiderivative at the upper and lower limits of integration and subtract the lower limit from the upper limit.

So, ( \int_{0}^{1} \frac{1}{1-x} , dx = -\ln|1-1| - (-\ln|1-0|) = -\ln|0| - (-\ln|1|) = -\ln|1| - (-\ln|1|) = 0 ).

Therefore, the value of the integral is ( 0 ).

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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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