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

The integral converges, and:

# int_0^oo \ (e^x)/(e^(2x)+1) \ dx = pi/4 #

Assuming a correction, let us first find:

Which is a standard integral, and so integrating:

And then reversing the earlier substitution:

If we look at the graph of the function, it would appear as if the bounded area is finite:

graph{(e^x)/(e^(2x)+1) [-10, 10, -2, 2]}

So let us test this prediction analytically. Consider the improper definite integral:

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To determine if the improper integral converges or diverges, we can use the limit comparison test. Let's compare the given integral to a simpler integral with which we're familiar.

Consider the integral of ( e^{-x} ) from 0 to infinity. This integral converges, as it represents an exponential decay function.

Now, let's compare ( \frac{e^x}{e^{2x} + 1} ) to ( e^{-x} ). As ( x ) approaches infinity, the term ( e^{-x} ) dominates ( \frac{e^x}{e^{2x} + 1} ).

So, we'll compare the given integral to ( \int_0^\infty e^{-x} ) using the limit comparison test.

[ \lim_{x \to \infty} \frac{\frac{e^x}{e^{2x} + 1}}{e^{-x}} = \lim_{x \to \infty} \frac{e^{2x}}{e^{2x} + 1} = 1 ]

Since the limit is a finite positive number, and since ( \int_0^\infty e^{-x} ) converges, by the limit comparison test, ( \int_0^\infty \frac{e^x}{e^{2x} + 1} dx ) also converges.

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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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