Could anyone help me with this monster?
it's the integral 1/((x^4) + 4) dx
Any help you could give would be Awesome!!!
$\displaystyle \int \frac{1}{x^4 + 4}dx = \int \frac{1}{(x^2 - 2x + 2)(x^2 + 2x + 2)}dx$
$\displaystyle \frac{Ax + C}{x^2 - 2x + 2} + \frac{Bx + D}{x^2 + 2x + 2}$
$\displaystyle Ax(x^2 + 2x + 2) + C(x^2 + 2x + 2) + Bx(x^2 - 2x + 2) + D(x^2 - 2x + 2) = 1$
Where C and D are 1/4,
$\displaystyle Ax(x^2 + 2x + 2) + 1/2x + Bx(x^2 - 2x + 2) = 0$
Well I just checked wolfram, the answer they've got is
$\displaystyle \frac{1}{8}tan^{-1}(x + 1) - \frac{1}{8}tan^{-1}(1 - x) + \frac{1}{16}ln(x^2 + 2x + 2) - \frac{1}{16}ln(x^2 - 2x + 2)$
Edit: Working backwards, that's
$\displaystyle \frac{1}{8} \int \frac{1}{1 + (x + 1)^2}dx - \frac{1}{8} \int \frac{-1}{1 + (1 - x)^2}dx$ $\displaystyle + \frac{1}{16} \int \frac{2x + 2}{x^2 + 2x + 2}dx - \frac{1}{16} \int \frac{2x - 2}{x^2 - 2x + 2}dx$
Edit: Where the denominators of the inverse tangents are just disguised forms of the other denominators.
$\displaystyle \frac{4}{x^4 + 4} = \frac{x^2 + 2}{x^4 + 4} - \frac{x^2 - 2}{x^4 + 4}=\frac{1 + \frac{2}{x^2}}{(x - \frac{2}{x})^2 + 4}- \frac{1- \frac{2}{x^2}}{(x + \frac{2}{x})^2 - 4}.$ let $\displaystyle y=x-\frac{2}{x}$ and $\displaystyle z=x + \frac{2}{x}.$ then: $\displaystyle 4 \int \frac{dx}{x^4 + 4} = \int \frac{dy}{y^2 + 4} - \int \frac{dz}{z^2 - 4},$ and the rest is easy now.