To solve the integral

$\int \frac{x}{\sqrt{x^2 + 1}} \, dx$

using the substitution $t = \sqrt{x^2 + 1}$, follow these steps:

Step 1: Substitution

Let: $t = \sqrt{x^2 + 1}$ Then: $t^2 = x^2 + 1$ Differentiating both sides with respect to $x$: $2t \frac{dt}{dx} = 2x$ $\frac{dt}{dx} = \frac{x}{t}$ $dx = \frac{t}{x} dt$

Step 2: Rewrite the integral

Substituting $t = \sqrt{x^2 + 1}$, we get: $\int \frac{x}{\sqrt{x^2 + 1}} \, dx = \int \frac{x}{t} \cdot \frac{t}{x} \, dt$ This simplifies to: $\int dt = t + C$

Step 3: Back-substitute

Since $t = \sqrt{x^2 + 1}$, the integral becomes: $\int \frac{x}{\sqrt{x^2 + 1}} \, dx = \sqrt{x^2 + 1} + C$

Thus, the solution to the integral is: $\sqrt{x^2 + 1} + C$

Would you like a further breakdown or have any questions?

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1. How do you apply substitution to other types of integrals?
2. What happens if you use different substitutions?
3. How do you verify the result of an integral?
4. Can you perform the same integral using trigonometric substitution?
5. What is the geometric interpretation of this integral?

Tip: Always check if a substitution simplifies both the integrand and the differential to make the integration process smoother.