Find the integral.
x^4/the square root of (x^5-1)
did you notice that the derivative of x^5=1 is 4x^4 which is a multiple of x^4, so it is straightforward
integral [ x4/(√(x^5 - 1)^(1/2)
= int [ x^4 (x^5 - 1)^(-1/2)
= (2/5) (x^5 - 1)^(1/2) + c
check by taking the derivative, it works
correctin: the derivative of x^5=1 is 5x^4
To find the integral of the function x^4 divided by the square root of (x^5 - 1), you can use a technique called u-substitution. Here's how you can do it step by step:
Step 1: Choose the substitution
Let u = x^5 - 1. We need to find du/dx to rewrite the function in terms of u.
Taking the derivative of both sides with respect to x, we get:
du/dx = 5x^4
Step 2: Rearrange the differential equation
Solving for dx, we get:
dx = du/(5x^4)
Step 3: Substitute into the integral
Rewrite the integral using u and dx substitutions:
∫ (x^4/sqrt(x^5 - 1)) dx = ∫ (x^4/sqrt(u)) (du/(5x^4))
Step 4: Simplify the integral
Cancel out the common terms of x^4:
∫ 1/(5√u) du
Step 5: Evaluate the integral
The integral of 1/(5√u) is (2/5)√u. Therefore, the antiderivative is:
(2/5)√u + C
Step 6: Replace u with the original variable x
Substituting back u = x^5 - 1:
(2/5)√(x^5 - 1) + C
So, the integral of x^4 divided by the square root of (x^5 - 1) is:
(2/5)√(x^5 - 1) + C, where C is the constant of integration.