help me integrate dx/(3x^3-5)^3
thanks
please don,t use that box? thing it always confuse me
no idea what box you're talking about. Probably an integral sign which does not render in your font.
Judging from the answer produced by wolframalpha.com
http://www.wolframalpha.com/input/?i=integral+dx%2F%283x^3-5%29^3
you have some major tricks to use.
To integrate the expression ∫ dx/(3x^3-5)^3, we can use a technique called substitution.
Here's how you can solve it step-by-step:
Step 1: Let's first simplify the denominator. We start by factoring out the constant 1/27 from the denominator:
(3x^3 - 5)^3 = [(∛(3x^3 - 5)]^3
Step 2: Now, let u = ∛(3x^3 - 5). Taking the derivative of both sides with respect to x, we have:
du/dx = d/dx [∛(3x^3 - 5)]
Step 3: Solve the above equation for dx:
dx = du / [(d/du) ∛(3x^3 - 5)]
dx = du / [(1/3) * (3x^2)] (using chain rule in the denominator)
Simplifying, it becomes:
dx = du / (3x^2)
Step 4: Now, substitute the values of u and dx into the integral:
∫ dx/(3x^3-5)^3 = ∫ du / (3x^2) * [(1/u)^3]
Simplifying further, we get:
∫ 1/(3x^2 * u^3) du
Step 5: Since u = ∛(3x^3 - 5), we can rewrite the expression as:
∫ 1/(3x^2 * (∛(3x^3 - 5))^3) du
Step 6: Now, inside the integral, make a substitution using u:
Let v = 3x^3 - 5. Taking the derivative of v with respect to x, we have:
dv/dx = 9x^2
Solving for dx, we get:
dx = dv / (9x^2)
Step 7: Substitute the values of v and dx into the integral:
∫ 1/(3x^2 * (∛v)^3) * (dv / (9x^2))
Step 8: Simplify the expression by canceling out x^2 and x^2 from the denominator:
∫ 1/(27 * v^3) dv
Step 9: Now, we can integrate:
∫ 1/(27 * v^3) dv = (1/27) ∫ v^(-3) dv
Using the power rule of integrals, we have:
(1/27) ∫ v^(-3) dv = (1/27) * (v^(-3+1) / (-3+1)) + C
(1/27) * (v^(-2) / (-2)) + C
Step 10: Substitute the value of v back in terms of x:
(1/27) * [(3x^3 - 5)^(-2) / (-2)] + C
Finally, simplify the expression if needed. Therefore, the solution is:
(1/27) * (3x^3 - 5)^(-2) / (-2) + C
This is the integral of dx/(3x^3-5)^3.