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

This would be integrated using partial fractions.

We can now write a system of equations:

The integral becomes:

These two integrals can be readily integrated.

Evaluate using the second fundamental theorem of calculus, which states that

Hopefully this helps!

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To evaluate the definite integral ∫(2x + 1)/(x + 1) from 0 to e - 1, you can first perform polynomial long division to rewrite the integrand as 2 - 1/(x + 1). Then integrate each term separately. The integral of 2 with respect to x from 0 to e - 1 is 2(e - 1). The integral of 1/(x + 1) from 0 to e - 1 can be found using the natural logarithm, resulting in ln(e) - ln(1) = 1 - 0 = 1. Therefore, the value of the definite integral is 2(e - 1) + 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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