How do you find the integral of #(1/(e^x+e^x))dx#?
Not entirely; it's just a small trick followed by substitution.
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To find the integral of ( \frac{1}{e^x + e^{x}} ,dx ), follow these steps:

Recognize that ( e^{x} ) can be rewritten as ( \frac{1}{e^x} ).
So, the integral becomes: [ \int \frac{1}{e^x + \frac{1}{e^x}} ,dx ]

Get a common denominator inside the fraction: [ \int \frac{1}{\frac{e^{2x} + 1}{e^x}} ,dx ]

Simplify the expression by inverting the fraction in the denominator: [ \int \frac{e^x}{e^{2x} + 1} ,dx ]

Use a substitution method to solve the integral. Let ( u = e^x ), which means ( du = e^x dx ). Notice that we already have ( e^x dx ) in our integral, so we can directly substitute.

Substitute ( u = e^x ) and ( du = e^x dx ) into the integral: [ \int \frac{du}{u^2 + 1} ]

This is a standard integral form, and its solution is known: [ \tan^{1}(u) + C ]

Substitute back for ( u = e^x ): [ \tan^{1}(e^x) + C ]
Thus, the integral of ( \frac{1}{e^x + e^{x}} ,dx ) is ( \tan^{1}(e^x) + C ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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