Integrate:( dx/root(x^2-4x)^3
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To integrate the given expression, dx / sqrt(x^2 - 4x)^3, we can use the substitution method.
Let's start by making a substitution:
Let u = x^2 - 4x.
Now, we need to find du/dx to express dx in terms of du.
Taking the derivative of both sides of our substitution equation with respect to x, we get:
du/dx = d/dx (x^2 - 4x)
Differentiating each term separately, we have:
du/dx = 2x - 4
Now, we can express dx in terms of du by rearranging the equation:
dx = du / (2x - 4)
Substituting the new expressions for dx and u back into the original integral, we have:
∫ (dx / sqrt(x^2 - 4x)^3) = ∫ (du / (2x - 4)) / sqrt(u)^3
Simplifying this expression, we get:
= (1/2) ∫ du / ((2x - 4)*sqrt(u)^3)
Now, we need to express x in terms of u. We can rearrange our original substitution equation to solve for x:
u = x^2 - 4x
=> u + 4x = x^2 - 4x + 4x
=> u + 4x = x^2
=> x^2 - 4x + u = 0
This is a quadratic equation, and we can use the quadratic formula to solve for x:
x = (4 ± sqrt(16 - 4u)) / 2
=> x = (4 ± sqrt(16 - 4u)) / 2
=> x = (4 ± 2sqrt(4 - u)) / 2
=> x = 2 ± sqrt(4 - u)
Now we can substitute this expression for x back into our integral:
(1/2) ∫ du / ((2x - 4)*sqrt(u)^3)
= (1/2) ∫ du / ((2(2 ± sqrt(4 - u)) - 4)*sqrt(u)^3)
= (1/2) ∫ du / (4 ± 2sqrt(4 - u) - 4)*sqrt(u)^3
= (1/2) ∫ du / (2sqrt(4 - u)*sqrt(u)^3)
= (1/2) ∫ du / (2sqrt(u^3(4 - u)))
Simplifying further, we have:
= (1/2) ∫ du / (2sqrt(u^4 - u^2))
Now, using the substitution method again, we can let v = u^2:
dv/du = 2u
=> du = dv / (2u)
Substituting this back into the integral, we get:
= (1/2) ∫ (dv / (2sqrt(v - u^2))) / (2u)
= (1/4) ∫ (dv / (u*sqrt(v - u^2)))
Integrating this integral with respect to v, we get:
= (1/4) * ln|v + sqrt(v - u^2)| + C
Finally, replacing v with its original substitution, u^2, we have:
= (1/4) * ln|u^2 + sqrt(u^2 - u^2)| + C
= (1/4) * ln|u^2| + C
= (1/4) ln| x^2 - 4x | + C
Therefore, the integral of dx / sqrt(x^2 - 4x)^3 is (1/4) ln| x^2 - 4x | + C.