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.
Best Answer
$$\frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i}=-\frac{1}{4} (-\frac{x}{2})\cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i} \,=-\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,$$ $$\int -\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,dx=\frac{1}{2}\sum_{i=0}^{\infty}\frac{1}{3i+2} \left(-\frac{x}{2}\right)^{3i+2} \,$$