I want to find the following series.

$$\sum\_{n\ge1} \frac{1}{4n^3-n}$$ 

So, fisrt of all, patial fraction : $$\frac{1}{4n^3-n}=\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{1}{n}.$$

Next, I consider the geometric series $$\sum\_{n\ge1} x^{2n-2}+x^{2n}-x^{n-1}=\frac{1}{1-x^2}+\frac{x^2}{1-x^2}-\frac{1}{1-x}\tag{1}$$ where ${-1\lt x\lt1}$.

Then, I integrate both side of (1) from $0$ to $1$:

$$\sum\_{n\ge1}\frac{1}{2n-1}+\frac{1}{2n+1}-\frac{1}{n}=\int\_0^1 \frac{x^2-x}{1-x^2}dx$$

Finally, I get the value that is equal to $ln(2)-1$ by improper integral but it's less than $0$. What did I do wrong? All help is appreciated.