Evaluating the integral $\int\frac{x-1}{x+4x^3}\mathrm dx$

Question

I need to solve this integral but I have no idea about how to procede, this is the integral:

$$\int \frac{x-1}{x+4x^3}\mathrm dx$$

This is how I solve the first part:

$$\int \frac{x}{x+4x^3}\mathrm dx - \int \frac{1}{x+4x^3}\mathrm dx$$

$$\int \frac{1}{1+4x^2}\mathrm dx - \int \frac{1}{x+4x^3}\mathrm dx$$

So I solved the first integral:

$$\int \frac{1}{1 + (2x)^2}\mathrm dx = \frac{1}{2}\arctan(2x) + C$$

But how can I solve the second?

$$- \int \frac{1}{x(1+4x^2)}\mathrm dx$$

Answer 1

Hint

Find $A;B,C$ such that

$$\frac{1}{x(1+4x^2)}=\frac{A}{x}+\frac{Bx+C}{1+4x^2}.$$

Answer 2

Hint: attack it with partial fractions: $$\frac{x-1}{x+4x^3} = \frac{x-1}{x(1+4x^2)} = \frac{A}{x}+\frac{Bx+C}{1+4x^2}.$$