The image contains a mathematical query regarding the Maclaurin series expansion of the function $\ln(1+x)$.

The Maclaurin series for $\ln(1+x)$ is derived from its Taylor series centered at $x = 0$. The general form for the Maclaurin series of a function $f(x)$ is:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \dots$

To find the series for $\ln(1+x)$, we need to compute its derivatives at $x = 0$.

1. The function itself at $x = 0$ is: $f(0) = \ln(1+0) = \ln(1) = 0$

2. The first derivative of $\ln(1+x)$ is: $f'(x) = \frac{1}{1+x}$ At $x = 0$, $f'(0) = \frac{1}{1+0} = 1$.

3. The second derivative is: $f''(x) = -\frac{1}{(1+x)^2}$ At $x = 0$, $f''(0) = -1$.

4. The third derivative is: $f^{(3)}(x) = \frac{2}{(1+x)^3}$ At $x = 0$, $f^{(3)}(0) = 2$.

5. The fourth derivative is: $f^{(4)}(x) = -\frac{6}{(1+x)^4}$ At $x = 0$, $f^{(4)}(0) = -6$.

Thus, the Maclaurin series for $\ln(1+x)$ is:

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$

In general, the series is:

$\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}, \quad \text{for} \quad |x| < 1$

Would you like further explanation or additional steps on this topic? Let me know if you need anything else!

Here are some related questions:

1. How do we determine the radius of convergence for the Maclaurin series of $\ln(1+x)$?
2. What are the general steps for deriving a Maclaurin series for other functions?
3. How can we approximate the value of $\ln(1+x)$ for small values of $x$ using its series?
4. Can the Maclaurin series for $\ln(1+x)$ be applied to complex numbers?
5. What is the significance of the alternating signs in the series for $\ln(1+x)$?

Tip: The convergence of the Maclaurin series for $\ln(1+x)$ is valid for $|x| < 1$. For values outside this range, the series does not converge.