# Convert integral to a series

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.

• @avid19 I know that $\int x^n \, dx = \frac{x^{n+1}}{n+1}$. But, how can I apply this to this product of $x \cdot \sum_{i=0}^{\infty} \left( - \frac{x}{2} \right)^{3i}$. Nov 22, 2015 at 18:50
• Just simplify the expression to $\sum \frac{(-1)^{3i}}{2^{3i}} x^{3i+1}$. Then you just have a sum of something that looks like $x^n$. Another hint, what is the integral of a sum?
– user223391
Nov 22, 2015 at 18:52
• You might need to be careful interchanging infinite sums and integrals because in general you can't. However in most case you'll deal with, it's okay and here it should come out okay.
– user223391
Nov 22, 2015 at 18:58

$$\frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i}=-\frac{1}{4} (-\frac{x}{2})\cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i} \,=-\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,$$ $$\int -\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,dx=\frac{1}{2}\sum_{i=0}^{\infty}\frac{1}{3i+2} \left(-\frac{x}{2}\right)^{3i+2} \,$$
$$\int \frac{x}{8+x^3}dx$$ $$=\lim_{h\to 0} \,\ h\sum_{r=1}^{n} \frac{rh}{8+(rh)^3}$$ $$=\lim_{n\to\infty} \,\ \frac{1}{n}\sum_{r=1}^{n} \frac{\frac{r}{n}}{8+(\frac{r}{n})^3}$$ $$=\lim_{n\to\infty} \,\ \sum_{r=1}^{n} \frac{rn}{8n^3+r^3}$$
• @SchrodingersCat In the second line, the summation's upper bound $n$ has not been defined Nov 18, 2020 at 20:14