# How do you find the antiderivative of #(e^(2x)+e^x)/(e^(2x)+1)#?

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To find the antiderivative of (\frac{e^{2x}+e^x}{e^{2x}+1}), you can use the method of partial fraction decomposition. First, rewrite the expression as (\frac{(e^x)^2+e^x}{(e^x)^2+1}). Then, factor the denominator as ((e^x)^2+1 = (e^x)^2 + 1^2). Now, we have:

(\frac{e^{2x}+e^x}{e^{2x}+1} = \frac{(e^x)^2+e^x}{(e^x)^2+1} = \frac{(e^x)(e^x+1)}{(e^x+1)(e^x-1)}).

Next, cancel out the common factor of ((e^x+1)):

(\frac{e^{2x}+e^x}{e^{2x}+1} = \frac{e^x}{e^x-1}).

Now, we can decompose this expression into partial fractions. Let (A) and (B) be constants:

(\frac{e^x}{e^x-1} = \frac{A}{e^x-1} + \frac{B}{e^x}).

Multiplying both sides by the denominator, we get:

(e^x = A(e^x) + B(e^x-1)).

Comparing coefficients of like terms, we find:

(1 = A + B), and (0 = -A).

Solving these equations, we find (A = 0) and (B = 1). Therefore, the partial fraction decomposition is:

(\frac{e^x}{e^x-1} = \frac{1}{e^x} + \frac{0}{e^x-1}).

Now, integrating each term separately:

(\int \frac{e^x}{e^x-1} , dx = \int \left(\frac{1}{e^x} + \frac{0}{e^x-1}\right) , dx).

This simplifies to:

(\ln|e^x| + C = \ln|e^x| + C),

where (C) is the constant of integration. Thus, the antiderivative of (\frac{e^{2x}+e^x}{e^{2x}+1}) is (\ln|e^x| + 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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