How do you find the antiderivative of #e^(2x) / (3+e^(2x))#?
easiest way always is to recognise the patterns
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you can plough through a whole series of subs but seeing the pattern is a real life saver.
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To find the antiderivative of ( \frac{e^{2x}}{3 + e^{2x}} ), perform a substitution to simplify the expression.
Let ( u = e^{2x} ).
Then, ( du = 2e^{2x} dx ).
Now, rewrite the integral using the substitution:
[ \int \frac{e^{2x}}{3 + e^{2x}} dx = \int \frac{1}{3 + u} \cdot \frac{du}{2} ]
This integral is now in a form that can be solved using a simple u-substitution:
[ \int \frac{1}{3 + u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{du}{3 + u} ]
Integrate with respect to u:
[ \frac{1}{2} \ln|3 + u| + C ]
Substitute back for u:
[ \frac{1}{2} \ln|3 + e^{2x}| + 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.
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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