Just to offer an alternative solution, let
[tex]f(x) = x + x(1-x^2) + x(1-x^2)^2 + \cdots = \displaystyle \sum_{n=0}^\infty x(1-x^2)^n[/tex]
Let F(x) denote the antiderivative of f(x). Then by the fundamental theorem of calculus, we can write, for instance,
[tex]F(x) = f(1) + \displaystyle \int_1^x f(t)\,\mathrm dt[/tex]
When x = 1, all terms in the sum corresponding to n ≥ 1 vanish, so that f (1) = 1.
Integrating the series and interchanging the sum and integral (terms and conditions apply - see Fubini's theorem) gives
[tex]\displaystyle \int f(x)\,\mathrm dx = \int \sum_{n=0}^\infty x(1-x^2)^n\,\mathrm dx = C + \sum_{n=0}^\infty \int x(1-x^2)^n\,\mathrm dx[/tex]
The constant C here corresponds exactly to f (1).
In the integral, substitute
[tex]y = 1-x^2 \implies \mathrm dy = -2x\,\mathrm dx[/tex]
so that it transforms and reduces to
[tex]\displaystyle -\frac12 \int y^n\,\mathrm dy = -\frac1{2(n+1)} y^{n+1} = -\frac1{2(n+1)}(1-x^2)^n[/tex]
Then we have
[tex]F(x) = 1 - \displaystyle \frac12 \sum_{n=0}^\infty \frac{(1-x^2)^{n+1}}{n+1}[/tex]
and by shifting the index to make the sum start at n = 1,
[tex]F(x) = 1 - \displaystyle \frac12 \sum_{n=1}^\infty \frac{(1-x^2)^n}n[/tex]
or equivalently,
[tex]F(x) = 1 - \displaystyle \frac12 \sum_{n=1}^\infty \frac{(-(x^2-1))^n}n[/tex]
Recall the Maclaurin expansion for ln(x) centered at x = 1, valid for |x - 1| < 1 :
[tex]\ln(x) = \displaystyle -\sum_{n=1}^\infty \frac{(-(x - 1))^n}n[/tex]
By comparing to this series, we observe that the series in F(x) converges to
[tex]\displaystyle -\sum_{n=1}^\infty \frac{(-(x^2-1))^n}n = \ln(x^2)[/tex]
so long as |x ² - 1| < 1. Then
[tex]F(x) = 1 + \dfrac12 \ln(x^2) = 1 + \ln(x)[/tex]
Differentiate both sides to recover f(x) = 1/x .
