Math Problem Statement

Evaluate the integral \( \int \frac{1}{e^x + e^{-x}} \, dx \)

Solution

To solve the integral

1ex+exdx\int \frac{1}{e^x + e^{-x}} \, dx

we can simplify the expression by rewriting the integrand. Notice that:

ex+ex=e2x+1ex.e^x + e^{-x} = \frac{e^{2x} + 1}{e^x}.

Thus, we can rewrite the integral as:

1ex+exdx=exe2x+1dx.\int \frac{1}{e^x + e^{-x}} \, dx = \int \frac{e^x}{e^{2x} + 1} \, dx.

Now, let's make a substitution to simplify further. Let:

u=exdu=exdxdx=duex=duu.u = e^x \Rightarrow du = e^x \, dx \Rightarrow dx = \frac{du}{e^x} = \frac{du}{u}.

Substituting ex=ue^x = u into the integral, we get:

exe2x+1dx=uu2+1duu=1u2+1du.\int \frac{e^x}{e^{2x} + 1} \, dx = \int \frac{u}{u^2 + 1} \cdot \frac{du}{u} = \int \frac{1}{u^2 + 1} \, du.

This integral is now in a standard form:

1u2+1du=arctan(u)+C.\int \frac{1}{u^2 + 1} \, du = \arctan(u) + C.

Finally, substitute back u=exu = e^x to get the answer:

1ex+exdx=arctan(ex)+C.\int \frac{1}{e^x + e^{-x}} \, dx = \arctan(e^x) + C.

The solution to the integral is:

arctan(ex)+C.\boxed{\arctan(e^x) + C}.

Would you like more details on any step or have any questions?

Here are some related questions for further exploration:

  1. How would you integrate 1exexdx\int \frac{1}{e^x - e^{-x}} \, dx?
  2. What is the derivative of arctan(ex)\arctan(e^x)?
  3. Can you evaluate xex+exdx\int \frac{x}{e^x + e^{-x}} \, dx?
  4. How do you simplify expressions involving hyperbolic functions like cosh(x)\cosh(x) and sinh(x)\sinh(x)?
  5. What are the properties of inverse trigonometric functions in integration?

Tip: When you encounter an integrand with exponential terms like ex+exe^x + e^{-x}, consider using substitutions or rewriting in terms of hyperbolic functions to simplify.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Substitution
Inverse Trigonometric Functions

Formulas

\( e^x + e^{-x} = \frac{e^{2x} + 1}{e^x} \)
Substitution: \( u = e^x \Rightarrow du = e^x \, dx \)
Integral of \( \frac{1}{u^2 + 1} \): \( \arctan(u) + C \)

Theorems

Substitution Rule for Integrals

Suitable Grade Level

College Level