1. **State the problem:** Find the first four non-zero terms of the Maclaurin series for the function $f(x) = \ln(1+x)$. 2. **Recall the Maclaurin series definition:** The Maclaurin series of a function $f(x)$ is given by $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \cdots$$ 3. **Calculate derivatives:** - $f(x) = \ln(1+x)$ - $f'(x) = \frac{1}{1+x}$ - $f''(x) = -\frac{1}{(1+x)^2}$ - $f'''(x) = \frac{2}{(1+x)^3}$ - $f^{(4)}(x) = -\frac{6}{(1+x)^4}$ 4. **Evaluate derivatives at $x=0$:** - $f(0) = \ln(1+0) = 0$ - $f'(0) = \frac{1}{1+0} = 1$ - $f''(0) = -1$ - $f'''(0) = 2$ - $f^{(4)}(0) = -6$ 5. **Substitute into the Maclaurin series formula:** $$f(x) = 0 + 1 \cdot x + \frac{-1}{2!}x^2 + \frac{2}{3!}x^3 + \frac{-6}{4!}x^4 + \cdots$$ 6. **Simplify factorials:** - $2! = 2$ - $3! = 6$ - $4! = 24$ 7. **Write the first four non-zero terms:** $$f(x) = x - \frac{x^2}{2} + \frac{2}{6}x^3 - \frac{6}{24}x^4 + \cdots = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$$ **Final answer:** The first four non-zero terms of the Maclaurin series for $\ln(1+x)$ are $$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$$