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.
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!
4 answers
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?
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.
I mean, I did all the algebra for you, and explained the 2nd example to show why it worked there.
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!
Thanks so much for the help!
4 answers