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

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.
  • #1
cmab
32
0
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? !
 
Physics news on Phys.org
  • #2
are you trying to do:
[tex]\int\frac{dx}{e^x+e^{-x}}[/tex]

if so, for a=b=c:
[tex]\int\frac{dx}{e^{ax}+e^{-bx}}=\int\frac{dx}{e^{cx}+e^{-cx}}=\frac{tan^{-1}(e^{cx})}{c}+C[/tex]
i hope that helps
 
  • #3
p53ud0 dr34m5 said:
are you trying to do:
[tex]\int\frac{dx}{e^x+e^{-x}}[/tex]

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


I don't get it :confused:
 
  • #4
[tex] \frac{1}{e^x + e^{-x}} = \frac{1}{\frac{(e^x)^2+1}{e^x}}[/tex]

Can you go from there?
 
  • #5
This might help you:

[tex]\frac{e^x + e^{-x}}{2} = \cosh x[/tex]
 

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.

4. How do you check your answer?

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.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
210
  • Introductory Physics Homework Help
Replies
19
Views
615
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
27
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
611
  • Introductory Physics Homework Help
Replies
12
Views
859
  • Introductory Physics Homework Help
Replies
9
Views
609
  • Introductory Physics Homework Help
Replies
6
Views
207
Back
Top