Calculate that intergal or disintegrate

x^3/sqrt(1-x^4).
Sometimes I wonder how would SAW movie look like, if it was replaced by math problems.

-1/4 ln |1 - x^4|

@tpr said ^

-1/4 ln |1 - x^4|

You're sure?

Dis integrand is f(x) = (x^3) / (√(1 – x^4))
Substitute u = 1 – x^4
du = – 4x^3 dx
dx = – du / (4x^3)

∫ f(x) dx =
= – ∫ (x^3) / (4x^3 sqrt(u)) du
= – (1/4) ∫ 1 / (sqrt(u)) du
= – (1/4) ∫ u^(–1/2) du
= – (1/4) 2u^(1/2) + C
= – (1/2) u^(1/2) + C
= – (1/2) √(1 – x^4) + C

For example we can use this result to calculate the following improper integrals:
∫_{-1}^{1} f(x) dx = 0 (symmetry of x^3)
∫_{0}^{1} f(x) dx = 1/2

Or using (3/4)^(1/4) ≈ 0.931 as the upper boundary:
∫_{0}^{(3/4)^(1/4)} (x^3) / (√(1 – x^4)) dx =
= – (1/2) √(1 – 3/4) – (– (1/2) √(1 – 0))
= – (1/2) √(1/4) – (– (1/2) )
= – (1/4) + (1/2)
= 1/4

However, assuming x to be a real number, special care must be taken when integrating that function over an interval other than
J = (-1, 1)
or any smaller interval contained within it. That's because f(x) only takes on real values on J. It's purely imaginary everywhere else along the real number-line. At x = ± 1 it goes to ± ∞ (coming from the inside of the interval) and -+ ∞ coming from the outside. Which is to say that the two-sided limits do not exist.

Shout out to @ResurgenceOfVoid:

I must admit that I do like the concept of an intergal.

"You're sure?"

• No, I forgot about the sqrt and gave the answer without the sqrt.

The answer is - sqrt (1 - x^4) / 2

I beg to differentiate...

@tpr said ^

"You're sure?"

• No, I forgot about the sqrt and gave the answer without the sqrt.

The answer is - sqrt (1 - x^4) / 2

You also forgot the +C

@benseshi said ^

You also forgot the +C

:nerd: