Let's solve the integrals numbered (4) and (5) from the image.

### Integral (4): \(\int \ln(1 + x^2) \, dx\)

This integral also requires integration by parts. We proceed similarly to how we approached the previous problem.

Let:
- \( u = \ln(1 + x^2) \), so \( du = \frac{2x}{1 + x^2} \, dx \)
- \( dv = dx \), hence \( v = x \)

Applying the integration by parts formula:

\[
\int \ln(1 + x^2) \, dx = x \ln(1 + x^2) - \int \frac{2x^2}{1 + x^2} \, dx
\]

Notice that:

\[
\frac{2x^2}{1 + x^2} = 2 - \frac{2}{1 + x^2}
\]

So, we can split the integral as follows:

\[
\int \ln(1 + x^2) \, dx = x \ln(1 + x^2) - \int \left(2 - \frac{2}{1 + x^2}\right) dx
\]

Now integrate the terms:

\[
\int 2 \, dx = 2x
\]

\[
\int \frac{2}{1 + x^2} \, dx = 2 \arctan(x)
\]

Therefore, the integral becomes:

\[
\int \ln(1 + x^2) \, dx = x \ln(1 + x^2) - 2x + 2\arctan(x) + C
\]

Thus, the final answer is:

\[
\boxed{x \ln(1 + x^2) - 2x + 2\arctan(x) + C}
\]

---

### Integral (5): \(\int \sqrt{1 + x} \, dx\)

To solve this integral, we can use a substitution method. Let:

\[
u = 1 + x, \quad \text{hence} \quad du = dx
\]

So, the integral becomes:

\[
\int \sqrt{u} \, du
\]

Now integrate \(\sqrt{u}\):

\[
\int u^{1/2} \, du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3} u^{3/2}
\]

Substituting back \( u = 1 + x \), we get:

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

Thus, the final answer is:

\[
\boxed{\frac{2}{3} (1 + x)^{3/2} + C}
\]

These are the solutions to the integrals numbered (4) and (5). Let me know if you need further clarification or if you have any other questions!

Here are 8 related questions you might find interesting:
1. How can integration by parts be applied to trigonometric functions?
2. What is the best strategy for selecting \(u\) and \(dv\) in integration by parts?
3. How can we evaluate integrals involving products of logarithmic and polynomial functions?
4. What are some common substitutions used to simplify integrals?
5. How can you integrate functions involving square roots of quadratic expressions?
6. What is the relationship between definite integrals and area under curves?
7. How can substitution be combined with other methods, like parts, in more complex integrals?
8. Can you explain the geometric interpretation of the integral \(\int \sqrt{1 + x} \, dx\)?

**Tip:** For integrals involving square roots, always consider substitution methods that simplify the square root, often by choosing a substitution that makes the expression under the square root a simple function of the new variable.

Learn how to solve the integrals \(\int \ln(1 + x^2) \, dx\) and \(\int \sqrt{1 + x} \, dx\) with step-by-step solutions. This problem covers key concepts in integral calculus, including the substitution method and integration by parts. Suitable for advanced high school students and above.

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