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

I found:

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We have that

Hence

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The answer is

Therefore,

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To evaluate the definite integral (\int_{0}^{1} \frac{x}{x+1} , dx), we can use a method called substitution.

Let (u = x + 1), then (du = dx).

When (x = 0), (u = 0 + 1 = 1).

When (x = 1), (u = 1 + 1 = 2).

Now, the integral becomes:

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

Expanding the fraction, we have:

[\int_{1}^{2} \left(1 - \frac{1}{u}\right) , du]

Now, integrate term by term:

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

[\left[u\right]*{1}^{2} - \left[\ln|u|\right]*{1}^{2}]

[= (2 - 1) - (\ln|2| - \ln|1|)]

[= 1 - (\ln(2) - 0)]

[= 1 - \ln(2)]

Therefore, (\int_{0}^{1} \frac{x}{x+1} , dx = 1 - \ln(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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