# How do you determine if the improper integral converges or diverges #int (1 / (u^2 + 3))du # from 0 to infinity?

The Improper Integral I converges to

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To determine if the improper integral (\int_{0}^{\infty} \frac{1}{{u^2 + 3}} , du) converges or diverges, you can use the limit comparison test. Compare it with a known integral that converges or diverges. In this case, consider comparing it with (\int_{0}^{\infty} \frac{1}{u^2} , du). Calculate the limit as (u) approaches infinity of the ratio of the two integrands. If the limit is finite and non-zero, both integrals converge or diverge together. If it's zero or infinite, they converge or diverge differently. Therefore, calculate:

[ \lim_{{u \to \infty}} \frac{{\frac{1}{{u^2 + 3}}}}{{\frac{1}{{u^2}}}} ]

If this limit is finite and non-zero, both integrals behave similarly. If it's zero or infinite, their behaviors differ.

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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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