To evaluate the integral

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

we can use the following method:

Step 1: Substitution

Let $u = 2 + x^2$, so that $du = 2x \, dx$. Thus, $x \, dx = \frac{du}{2}$, and we rewrite the integral in terms of $u$:

$\int \frac{x^2 + 1}{\sqrt{2 + x^2}} \, dx = \int \frac{(u - 2) + 1}{\sqrt{u}} \cdot \frac{du}{2x}$

Since $u = 2 + x^2$, we also know $x^2 = u - 2$. Therefore:

$\int \frac{(u - 2) + 1}{\sqrt{u}} \cdot \frac{du}{2x} = \int \frac{u - 1}{\sqrt{u}} \cdot \frac{du}{2x}$

Step 2: Simplification

Now, simplify the integral:

$\frac{1}{2} \int \left( \frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}} \right) du = \frac{1}{2} \int \left( \sqrt{u} - \frac{1}{\sqrt{u}} \right) du$

Step 3: Integration

Now, integrate each term:

$\frac{1}{2} \left( \int \sqrt{u} \, du - \int \frac{1}{\sqrt{u}} \, du \right)$

The integral of $\sqrt{u}$ is $\frac{2}{3} u^{3/2}$, and the integral of $\frac{1}{\sqrt{u}}$ is $2\sqrt{u}$. Thus, we have:

$\frac{1}{2} \left( \frac{2}{3} u^{3/2} - 2\sqrt{u} \right) + C$

Step 4: Substitute back $u = 2 + x^2$

Finally, substitute $u = 2 + x^2$ back into the expression:

$\frac{1}{3} (2 + x^2)^{3/2} - \sqrt{2 + x^2} + C$

Final Answer

$\int \frac{x^2 + 1}{\sqrt{2 + x^2}} \, dx = \frac{1}{3} (2 + x^2)^{3/2} - \sqrt{2 + x^2} + C$

Would you like more details on any step? If you have further questions, feel free to ask!

Here are some related questions you might find interesting:

1. How do you solve integrals involving square roots in the denominator?
2. What are common substitutions used in integral calculus?
3. Can you explain the method of integration by parts with an example?
4. What are definite integrals, and how do they differ from indefinite integrals?
5. How can the chain rule be applied in the context of integration?

Tip: When dealing with integrals involving square roots, substitution is often a powerful technique to simplify the expression.