# How do you find the antiderivative of #[e^(2x)/(4+e^(4x))]#?

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To find the antiderivative of (\frac{e^{2x}}{4+e^{4x}}), you can use a substitution method. Let (u = e^{2x}), then (du = 2e^{2x} dx). Rearrange to get (dx = \frac{du}{2e^{2x}}). Substitute (u) and (dx) into the integral:

[ \int \frac{e^{2x}}{4+e^{4x}} dx = \int \frac{1}{4+u^2} \times \frac{du}{2u} ]

Now, this integral can be solved using the arctangent function:

[ \frac{1}{2} \int \frac{1}{1 + \left(\frac{u}{2}\right)^2} , du ]

Using the substitution (w = \frac{u}{2}), (dw = \frac{1}{2} du), the integral becomes:

[ \int \frac{1}{1 + w^2} , dw ]

Which is the standard integral for arctangent. Therefore, the antiderivative is:

[ \frac{1}{2} \arctan\left(\frac{u}{2}\right) + C ]

Finally, substitute back (u = e^{2x}) to get:

[ \frac{1}{2} \arctan\left(\frac{e^{2x}}{2}\right) + C ]

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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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