Areas and Alternating Series Reposted.

Let $|x| < 1$ and $(0 < a < 1)$ .

Let $f(x) = \sum_{n = 1}^{\infty} (-1)^{n + 1} \dfrac{x^n}{n}$ .

$\implies \dfrac{d}{dx}(f(x)) = \sum_{n = 1}^{\infty} (-1)^{n + 1} x^{n - 1} = \dfrac{1}{1 + x} \implies f(x) = \int_{0}^{x} \dfrac{1}{1 + x} dx = \ln(1 + x)$ .

For $\int_{0}^{a} \ln(1 + x) dx$

$\int_{0}^{a} \ln(1 + x) dx = (x\ln(1 + x) + \ln(1 + x) - x)|_{0}^{a} = (x + 1)\ln(1 + x) - x|_{0}^{a} = (a + 1)\ln(a + 1) - a$ .

$\sum_{n = 1}^{\infty} n x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2}$

$\implies g(x) = \sum_{n = 1}^{\infty} (-1)^{n + 1} n x^n = - \sum_{n = 1}^{\infty} n (-x)^n = \dfrac{x}{(1 + x)^2}$ .

Let $u = 1 + x \implies du = dx \implies \int_{0}^{a} g(x) dx = \int_{1}^{a + 1} \dfrac{1}{u} - \dfrac{1}{u^2} du = (\ln(u) + \dfrac{1}{u})|_{0}^{a + 1} = \ln(a + 1) + \dfrac{1}{a + 1} - 1$

$\implies \int_{0}^{a} f(x) - g(x) dx = a\ln(a + 1) - \dfrac{a^2}{a + 1} = a\ln(a + 1) - (2a - 1)$ $\implies a^2 + a - 1 = 0 \implies a = \dfrac{-1 \pm \sqrt{5}}{2}$ $(0 < a < 1) \implies a = \dfrac{\sqrt{5} - 1}{2} \implies$ $\int_{0}^{\frac{\sqrt{5} - 1}{2}} f(x) - g(x) dx =$ $\dfrac{\sqrt{5} - 1}{2}\ln(\dfrac{1 + \sqrt{5}}{2}) - \dfrac{3 - \sqrt{5}}{\sqrt{5} + 1} \approx \boxed{0.061337}$ .