# How do you find the antiderivative of # {(e^x)/ [(e^(2x)) - 1]}#?

Plugging these in:

This is a fairly common integral:

So:

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To find the antiderivative of ( \frac{e^x}{e^{2x}-1} ), you can use the technique of substitution. Let ( u = e^x ), then ( du = e^x dx ). Substitute these into the integral:

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

Now, rewrite the integrand:

[ \frac{1}{u^2 - 1} = \frac{1}{(u-1)(u+1)} ]

Use partial fraction decomposition to break it down:

[ \frac{1}{(u-1)(u+1)} = \frac{A}{u-1} + \frac{B}{u+1} ]

Solve for ( A ) and ( B ) by finding a common denominator:

[ 1 = A(u+1) + B(u-1) ]

[ 1 = Au + A + Bu - B ]

[ 1 = (A+B)u + (A-B) ]

Matching coefficients, ( A + B = 0 ) and ( A - B = 1 ), thus ( A = \frac{1}{2} ) and ( B = -\frac{1}{2} ).

Now, integrate each term:

[ \int \frac{1}{(u-1)(u+1)} du = \frac{1}{2} \ln|u-1| - \frac{1}{2} \ln|u+1| + C ]

Replace ( u ) with ( e^x ):

[ \frac{1}{2} \ln|e^x-1| - \frac{1}{2} \ln|e^x+1| + C ]

That is the antiderivative of ( \frac{e^x}{e^{2x}-1} ).

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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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