1. ## tough integral

Could anyone help me with this monster?

it's the integral 1/((x^4) + 4) dx

2. try a u substitution. that is, let u = (x^4) + 4 and then take the intergral of 1/u

3. Originally Posted by o&apartyrock
try a u substitution. that is, let u = (x^4) + 4 and then take the intergral of 1/u
Not possible since $du=4x^{3}dx$, and there is no $4x^{3}$ term in the integral.

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

$\frac{Ax + C}{x^2 - 2x + 2} + \frac{Bx + D}{x^2 + 2x + 2}$

$Ax(x^2 + 2x + 2) + C(x^2 + 2x + 2) + Bx(x^2 - 2x + 2) + D(x^2 - 2x + 2) = 1$

Where C and D are 1/4,

$Ax(x^2 + 2x + 2) + 1/2x + Bx(x^2 - 2x + 2) = 0$

5. ## tried that

I tried messing with a u sub but didn't get anywhere.

My prof said that this problem uses most of the techniques, as far as integral calc goes. He then went on to say that once you get past clac 2, this problem is easy.

Thanks prof! lol

6. ## aahhh

Originally Posted by derfleurer
$\int \frac{1}{x^4 + 4}dx = \int \frac{1}{(x^2 - 2x + 2)(x^2 + 2x + 2)}dx$

$\frac{Ax + C}{x^2 - 2x + 2} + \frac{Bx + D}{x^2 + 2x + 2}$

$Ax(x^2 + 2x + 2) + C(x^2 + 2x + 2) + Bx(x^2 - 2x + 2) + D(x^2 - 2x + 2) = 1$

Where C and D are 1/4,

$Ax(x^2 + 2x + 2) + 1/2x + Bx(x^2 - 2x + 2) = 0$
Hmmmm...

Thanks that helps me get going. This doesn't look like it's going to be fun though. hehehe.

7. Well I just checked wolfram, the answer they've got is

$\frac{1}{8}tan^{-1}(x + 1) - \frac{1}{8}tan^{-1}(1 - x) + \frac{1}{16}ln(x^2 + 2x + 2) - \frac{1}{16}ln(x^2 - 2x + 2)$

Edit: Working backwards, that's

$\frac{1}{8} \int \frac{1}{1 + (x + 1)^2}dx - \frac{1}{8} \int \frac{-1}{1 + (1 - x)^2}dx$ $+ \frac{1}{16} \int \frac{2x + 2}{x^2 + 2x + 2}dx - \frac{1}{16} \int \frac{2x - 2}{x^2 - 2x + 2}dx$

Edit: Where the denominators of the inverse tangents are just disguised forms of the other denominators.

8. Originally Posted by redepoch7
Could anyone help me with this monster?

it's the integral 1/((x^4) + 4) dx

$\frac{4}{x^4 + 4} = \frac{x^2 + 2}{x^4 + 4} - \frac{x^2 - 2}{x^4 + 4}=\frac{1 + \frac{2}{x^2}}{(x - \frac{2}{x})^2 + 4}- \frac{1- \frac{2}{x^2}}{(x + \frac{2}{x})^2 - 4}.$ let $y=x-\frac{2}{x}$ and $z=x + \frac{2}{x}.$ then: $4 \int \frac{dx}{x^4 + 4} = \int \frac{dy}{y^2 + 4} - \int \frac{dz}{z^2 - 4},$ and the rest is easy now.

9. Originally Posted by NonCommAlg
$\frac{4}{x^4 + 4} = \frac{x^2 + 2}{x^4 + 4} - \frac{x^2 - 2}{x^4 + 4}=\frac{1 + \frac{2}{x^2}}{(x - \frac{2}{x})^2 + 4}- \frac{1- \frac{2}{x^2}}{(x + \frac{2}{x})^2 - 4}.$ let $y=x-\frac{2}{x}$ and $z=x + \frac{2}{x}.$ then: $4 \int \frac{dx}{x^4 + 4} = \int \frac{dy}{y^2 + 4} - \int \frac{dz}{z^2 - 4},$ and the rest is easy now.
sorry but where did the 4 in the numerator of part one come from? Does it matter? Can you just factor that out?

ex. $\frac{4}{x^4 + 4}$

10. Originally Posted by redepoch7
sorry but where did the 4 in the numerator of part one come from? Does it matter? Can you just factor that out?

ex. $\frac{4}{x^4 + 4}$
Since you were dealing with $\frac{1}{x^4+1}$, follow NonCommAlg's suggestion, but then multiply your answer by $\frac{1}{4}$.