# What is the integral of #(x+1)/(x(x^2+x-6))#?

Well, for one, we can factor the denominator:

Now we just have linear denominators, which is a fairly straightforward decomposition.

Simplify the left side so that it looks like a general polynomial. First, distribute:

Group together terms.

This means we have a system of three equations:

Thus:

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To find the integral of ((x+1)/(x(x^2+x-6))), you first factor the denominator, then use partial fraction decomposition. Factoring the denominator gives (x(x^2+x-6) = x(x+3)(x-2)). The partial fraction decomposition of ((x+1)/(x(x^2+x-6))) is (A/x + B/(x+3) + C/(x-2)), where (A), (B), and (C) are constants. Solving for these constants yields (A = 1/2), (B = -1/4), and (C = 1/4). Therefore, the integral becomes ((1/2)\ln|x| - (1/4)\ln|x+3| + (1/4)\ln|x-2| + C), where (C) is the constant of integration.

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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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