We are tasked with evaluating the integral:

$I = \int \frac{x e^{\sqrt{x^2 - 1}}}{\sqrt{x^2 - 1}} \, dx$

Step 1: Substitution

We will perform the substitution:

$u = \sqrt{x^2 - 1} \quad \text{so} \quad du = \frac{x}{\sqrt{x^2 - 1}} \, dx$

This substitution simplifies the integral as follows:

• $x^2 - 1 = u^2$, so $x = \sqrt{u^2 + 1}$
• $dx = \frac{du}{\sqrt{u^2 + 1}}$

Thus, the integral becomes:

$I = \int e^u \, du$

Step 2: Integration

The integral of $e^u$ with respect to $u$ is straightforward:

$I = e^u + C$

Step 3: Back-substitution

Now, we substitute back $u = \sqrt{x^2 - 1}$ to express the result in terms of $x$:

$I = e^{\sqrt{x^2 - 1}} + C$

Final Answer:

The result of the integral is:

$\boxed{e^{\sqrt{x^2 - 1}} + C}$

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Would you like more details on any part of the solution? Feel free to ask!

Here are 5 related questions to explore:

1. How do we handle integrals involving exponential functions and square roots?
2. What are some other types of substitution methods in calculus?
3. Can this method be applied to integrals with trigonometric functions?
4. How do we deal with inverse hyperbolic functions in integrals?
5. What are the basic rules of integration that help simplify complex integrals?

Tip: When encountering integrals with square roots in the denominator, a substitution involving the square root can often help simplify the expression.