Question

I am given two integrals a and b

a) ∫ 1/(1 + x^4) dx

b) ∫ x/(1 + x^4) dx

The main difference between the two integrals appears to be the "1" and "x" on the numerators. While they both resembles closely to the basic integration of arctan:

∫ du/(a^2 + u^2) = 1/a * arctan(u/a) + C

Can I substitute both a and b to follow this basic integration formula without breaking any rules? I worked out each one and got the same result as (1/2) * arctan(x^2) + C so I'm not completely sure if that would work for both a and b.

Any help is greatly appreciated!
Steve
Nope. Not that easy.

x^4+1 = x^4+2x^2+1 - 2x^2
= (x^2+1)^2 - (?2 x)^2
= (x^2+?2 x+1)(x^2-?2 x+1)

1/(x^4+1) = 1/[(x^2+?2 x+1)(x^2-?2 x+1)]
= 1/(2?2) ((x+?2)/(x^2+?2 x+1) - (x-?2)/(x^2-?2 x+1))

= 1/(2?2) ((2x+?2)/(x^2+?2 x+1) - x/(x^2+?2 x+1) - (2x-?2)/(x^2-?2 x+1) + x/(x^2-?2 x+1))

Now,
(2x+?2)/(x^2+?2 x+1) = du/u
(2x-?2)/(x^2-?2 x+1) = dv/v

x^2+?2 x+1 = (x + 1/?2)^2 + 1/2
x^2-?2 x+1 = (x - 1/?2)^2 + 1/2
On those, you can let use your arctan trick. In the end, you wind up with

http://www.wolframalpha.com/input/?i=%E2%88%AB+1%2F(1+%2B+x%5E4)+dx

For x/(x^4+1) it's a bit easier. Let u=x^2 and then you just have

1/2 du/(1+u^2)

and the arctan falls right out.
Ray
So, why is it that we can't use the easy trick method for the first one? Is it because of the "1" on the numerator that isn't a variable?
Steve
that is correct. There is no du to work with.

I mean, I did all the algebra for you, and explained the 2nd example to show why it worked there.
Ray
Yeah I see it, I just tend to overthink way too much than I should so I end up asking obvious questions. My apologies.

Thanks so much for the help!