# Find Indefinite Integral of 1/[(e^x)+(e^-x)]

• cmab
In summary, we can find the indefinite integral of 1/[(e^x)+(e^-x)] by using the formula \int\frac{dx}{e^x+e^{-x}}=\int\frac{dx}{e^{cx}+e^{-cx}}=\frac{tan^{-1}(e^{cx})}{c}+C, where c is the value of a, b, and c in the original equation. Additionally, we can use the formula \frac{e^x + e^{-x}}{2} = \cosh x to simplify the expression.
cmab
How do I find the indefinite integral of 1/[(e^x)+(e^-x)] ?

Do I have to multiply by [(e^x)-(e^-x)]/[(e^x)-(e^-x)] to eliminate the denominator? !

are you trying to do:
$$\int\frac{dx}{e^x+e^{-x}}$$

if so, for a=b=c:
$$\int\frac{dx}{e^{ax}+e^{-bx}}=\int\frac{dx}{e^{cx}+e^{-cx}}=\frac{tan^{-1}(e^{cx})}{c}+C$$
i hope that helps

p53ud0 dr34m5 said:
are you trying to do:
$$\int\frac{dx}{e^x+e^{-x}}$$

if so, for a=b=c:
$$\int\frac{dx}{e^{ax}+e^{-bx}}=\int\frac{dx}{e^{cx}+e^{-cx}}=\frac{tan^{-1}(e^{cx})}{c}+C$$
i hope that helps

I don't get it

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

Can you go from there?

$$\frac{e^x + e^{-x}}{2} = \cosh x$$

## 1. What is an indefinite integral?

An indefinite integral is a mathematical operation that finds the function whose derivative equals a given function. In other words, it is the reverse process of differentiation.

## 2. What is the given function in this integral?

The given function is 1/[(e^x)+(e^-x)], which can also be written as (e^-x)/(e^x + e^-x).

## 3. How do you solve this integral?

To solve this integral, we can use the substitution method. Let u = e^x + e^-x, then du = (e^x - e^-x)dx. Substituting this into the integral, we get ∫1/u du, which is equal to ln|u| + C. Therefore, the indefinite integral of 1/[(e^x)+(e^-x)] is ln|e^x + e^-x| + C.

To check the answer, we can differentiate ln|e^x + e^-x| + C with respect to x and see if we get 1/[(e^x)+(e^-x)]. The derivative of ln|e^x + e^-x| + C is (e^x - e^-x)/(e^x + e^-x) = (e^-x)/(e^x + e^-x), which is equal to the given function.

## 5. Can this integral be solved in any other way?

Yes, this integral can also be solved using the partial fraction decomposition method. By decomposing the given function into 1/e^x + 1/e^-x and then integrating each term separately, we can arrive at the same answer of ln|e^x + e^-x| + C.

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