# How do you find the indefinite integral of #int (x^2+2x+3)/(x^2+3x^2+9x)#?

So:

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To find the indefinite integral of (\int \frac{x^2+2x+3}{x^2+3x^2+9x} , dx), you first need to factor the denominator.

[x^2+3x^2+9x = x^2(1+3x+9)] [= x^2(3x^2+3x+3)] [= x^2(3)(x^2+x+1)]

So the integral becomes:

[\int \frac{x^2+2x+3}{x^2(3)(x^2+x+1)} , dx]

Now, we can use partial fraction decomposition. We have:

[\frac{x^2+2x+3}{x^2(3)(x^2+x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+x+1}]

Solving for (A), (B), (C), and (D), then integrating each term separately will yield the indefinite integral.

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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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