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!
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.
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.
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!
To solve the first integral ∫(1/(1 + x^4)) dx, you can make the following substitution:
Let u = x^2. Then, du = 2x dx.
Substituting these values into the integral, we get:
∫(1/(1 + x^4)) dx = ∫(1/(1 + u^2)) (1/2x) du
= (1/2) ∫(1/(1 + u^2)) du
Now, notice that the integrand is the same as the basic arctan form, with a^2 = 1 and u = x. Therefore, we can apply the formula:
∫(1/(1 + u^2)) du = arctan(u) + C
Substituting back the original variable:
(1/2) ∫(1/(1 + u^2)) du = (1/2) * arctan(u) + C
Finally, replacing u with x^2, we get:
(1/2) * arctan(x^2) + C
So, your result for the first integral is correct.
Now, let's move on to the second integral, ∫(x/(1 + x^4)) dx.
Again, we can make a substitution to simplify the integral:
Let u = x^2. Then, du = 2x dx.
Substituting these values into the integral, we get:
∫(x/(1 + x^4)) dx = ∫(u/(1 + u^2)) (1/2x) du
= (1/2) ∫(u/(1 + u^2)) du
Unlike the first integral, the second integral cannot be directly solved using the basic arctan form. However, we can express the numerator as a derivative of the denominator by applying the chain rule:
d/dx (arctan(u)) = 1/(1 + u^2) * du/dx
So, du/dx = (1 + u^2) * d/dx (arctan(u))
Therefore, we can rewrite the integral as:
(1/2) ∫(u/(1 + u^2)) du = (1/2) * [d/dx(arctan(u))] + C
Now, substituting back the original variable for u, we get:
(1/2) * [d/dx(arctan(x^2))] + C
So, the second integral can be written as the derivative of arctan(x^2) with respect to x.
In conclusion, you cannot directly apply the basic arctan formula to the second integral. However, by expressing the numerator as a derivative of the denominator, you can simplify the integral.