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

Answer 1

Evelyn Agard

The integral is not convergent

The integral is not convergent.

As the integrand is not continuous for #x=1# we can evaluate the improper integral:

#int_0^1 (dx)/(1-x) = lim_(t->1^-) int_0^t (dx)/(1-x)#

Now we have that:

# int_0^t (dx)/(1-x) =- int_0^t (d(1-x))/(1-x) = -[ln abs (1-x)]_0^t = - ln abs (1-t)#

and

#lim_(t->1^-) -ln(1-t) = +oo#

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Answer 2

Emma Aldridge

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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Answer from HIX Tutor

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.