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!

Answer 1

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.

Answer 2

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?

Answer 3

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.

Answer 4

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!