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Find a power series representation of $f(x)=\int_{0}^{x} \frac{1}{8-t^{3}} d t$ in powers of $\mathrm x.$

$\frac{1}{8-t^{3}}=\frac{1}{8[1-t^3 / 8]}=\frac{1}{8} \sum_{n=0}^{\infty}\left(\frac{t^{3}}{8}\right)^{n}=\frac{1}{8} \sum_{n=0}^{\infty} \frac{\left(t^{3}\right)^{n}}{8^{n}}=\frac{1}{8} \sum_{n=0}^{\infty} \frac{t^{3 n}}{8^{n}}$

$f(x)=\int_{0}^{x} \frac{1}{8-t^{3}} d t=\frac{1}{8} \int_{0}^{x}\left(\sum_{n=0}^{\infty} \frac{t^{3 n}}{8^{n}}\right) d t$

$=\frac{1}{8^1} \sum_{n=0}^{\infty} \frac{1}{8^{n}} \int_{0}^{x} t^{3 n} \ d t$

$=\sum_{n=0}^{\infty} \frac{1}{8^{n+1}}\left[\frac{t^{3 n+1}}{3 n+1}\right]_{0}^{x}$

$=\sum_{n=0}^{\infty} \frac{1}{8^{n+1}(3 n+1)}\left(x^{3 n+1}\right)$

Find a power series representation for $\ln (1+x)$ and its radius of convergence.

$\frac{d}{d x}[\ln (1+x)]=\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum(-x)^{n}=1-x+x^{2}-x^{3}\cdots$

$\ln (1+x)=\int \frac{1}{1+x} d x=\int 1-x+x^{2}-x^{3}-\ldots d x$

$=x+\frac{x^{3}}{3}-\frac{x^{4}}{4} \dots+C$

$=\int\left[\sum_{n=0}^{\infty}(-x)^{n}\right] d x=\sum_{n=0}^{\infty} \int(-x)^{n} d x$

$=\sum_{n=0}^{\infty} \int(-1)^{n} x^{n} d x=\sum_{n=0}^{\infty}(-1)^{n} \int x^{n} d x$

$=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n+1}}{n+1}+c, \quad|x|<1$

To find the value of $c \Rightarrow$ put $x=0$

$\ln (1+0)^{0}=c \quad \Rightarrow C=0$

$\ln (1+x)=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n+1}}{n+1} \quad, \quad|x|<1$

Radius of Convergence $=1$