# How do you evaluate the definite integral #int (x+2)/(x+1)# from #[0, e-1]#?

Both of these have common antiderivatives:

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To evaluate the definite integral (\int_0^{e-1} \frac{x+2}{x+1} , dx), we can use the technique of integration by substitution. Let (u = x + 1), then (du = dx).

Now, when (x = 0), (u = 0 + 1 = 1), and when (x = e - 1), (u = (e - 1) + 1 = e).

The integral becomes: [\int_1^e \frac{u + 1}{u} , du]

This simplifies to: [\int_1^e \left(1 + \frac{1}{u}\right) , du]

Now, integrate term by term: [= \left[u + \ln|u|\right]_1^e]

Substitute back (u = x + 1): [= \left[(x + 1) + \ln|x + 1|\right]_0^{e-1}]

[= ((e - 1) + \ln|e - 1|) - (0 + \ln|0 + 1|)]

[= (e - 1 + \ln|e - 1|) - \ln(1)]

[= e - 1 + \ln(e - 1)]

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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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