I have to find an infinitite series expansion for the integral: $$\int \frac{x}{8+x^3} \, dx$$
First, I started by determining the Taylor series of the integrand $$\frac{x}{8+x^3}=\frac{x}{8} \cdot \frac{1}{1-(-(x/2)^3)} = \frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i}$$
Then, I integrate $$\int \frac{x}{8+x^3} \, dx = -\frac{1}{8} \int x \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i} \, dx$$
But, I'm not sure how to continue.
Thank you for your help.
$\endgroup$ 32 Answers
$\begingroup$$$\frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i}=-\frac{1}{4} (-\frac{x}{2})\cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i} \,=-\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,$$ $$\int -\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,dx=\frac{1}{2}\sum_{i=0}^{\infty}\frac{1}{3i+2} \left(-\frac{x}{2}\right)^{3i+2} \,$$
$\endgroup$ 0 $\begingroup$$$\int \frac{x}{8+x^3}dx$$ $$=\lim_{h\to 0} \,\ h\sum_{r=1}^{n} \frac{rh}{8+(rh)^3}$$ $$=\lim_{n\to\infty} \,\ \frac{1}{n}\sum_{r=1}^{n} \frac{\frac{r}{n}}{8+(\frac{r}{n})^3}$$ $$=\lim_{n\to\infty} \,\ \sum_{r=1}^{n} \frac{rn}{8n^3+r^3}$$
Is this the series what you were looking for?
Please confirm.
