# Finding the power series representation for $\ln(1 -10x)$ via integration.

I'm trying to find the power series representation for $\ln(1-10x)$

Attempt at solution:

$$\ln(1-10x) = \int {-10\over1-10x} \ dx = -10 \int \sum_{n=0}^\infty (10x)^n dx$$ $$= -10 \sum_{n=0}^\infty {10^n x^{n+1}\over n+1} + C$$

$$C = \ln(1) = 0, \; \text {letting x = 0 to find value of C}$$

The answer is not being accepted, I think I might be making some mistakes with coefficients, but not sure how. I'm leaving $10^n$ since the integration is over $x$.

Note that $$\dfrac{1}{1 - 10x} = \sum_{n=0}^{+\infty}(10x)^n \quad \text{if} \quad |10x|< 1$$ integrating both sides we get, $$\int \dfrac{1}{1 - 10x}dx = \int \sum_{n=0}^{+\infty}(10x)^n = \sum_{n=0}^{+\infty} 10^n\int x^ndx = \sum_{n=0}^{+\infty} 10^n\dfrac{x^{n+1}}{n+1} + C$$ But, $$\int \dfrac{1}{1 - 10x}dx = -\dfrac{1}{10}\int \dfrac{(-10)dx}{1 - 10x} = \ln(1 - 10x)$$ So that $$\ln(1 - 10x) = -\sum_{n=0}^{+\infty} 10^{n+1}\dfrac{x^{n+1}}{n+1} + C$$ For $x = 0$, it follows that $C = 0$. Thus, $$\ln(1 - 10x) = -\sum_{n=0}^{+\infty} 10^{n+1}\dfrac{x^{n+1}}{n+1}$$ Note that x = 0 is contained in the interval of convergence. Therefore, its replacement is valid.