0

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?

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

rubik
  • 9,344
Christian Giupponi
  • 687
  • 2
    Perhaps a slightly more elegant approach would be to substitute $x=\dfrac1t$ first, and then separate the new integral into an arctangent and a natural logarithm. – Lucian Feb 06 '15 at 09:32

2 Answers2

2

Hint

Find $A;B,C$ such that

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

idm
  • 11,824
2

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

Ivo Terek
  • 77,665