integrate
(1)/(-4ln)3x)^3
-27x^2 cuberoot-3x^3-5
and using the graph y=squareroot of x which points are closest to (6,0).
thank you!
To integrate the expression, (1/(-4ln(3x))^3), we can use the power rule of integration and apply substitution. Here's the step-by-step process:
Step 1: Rewrite the expression in a simpler form:
1/(-4ln(3x))^3
Step 2: Let u = 3x. Express the expression in terms of u:
1/(-4ln(u))^3
Step 3: Apply the power rule of integration:
∫(1/(-4ln(u))^3) du
Step 4: Distribute the negative power:
∫(-4ln(u))^(-3) du
Step 5: Use the substitution, let v = -4ln(u):
du = (1/u) dv, rearrange to get:
dv = -u du
Step 6: Substitute the values of v and dv into the integral:
-∫v^(-3) (1/u) dv
Step 7: Simplify the expression and integrate with respect to v:
-∫v^(-4) dv
= -(1/(-3)) * v^(-3)
= 1/3v^3
Step 8: Replace v with the original variable (u) and simplify:
= 1/(3*(-4ln(u))^3)
Step 9: Undo the substitution by replacing u with the original variable (3x):
= 1/(3*(-4ln(3x))^3)
Therefore, the integral of (1/(-4ln(3x))^3) with respect to x is 1/(3*(-4ln(3x))^3).