Integrate x^2/rootx-1..
this one is a bit less complicated than the last one.
∫x^2/√(x-1)
Let u^2 = x-1
2u du = dx
x = 1+u^2
and we have
∫(1+u^2)^2/u 2u du
2∫(1+u^2)^2 du
2∫1 + 2u^2 + u^4 du
2(u + 2/3 u^3 + 1/5 u^5)
2/15 √(x-1)(15 + 10(x-1) + 3(x-1)^2)
2/15 √(x-1)(15 + 10x - 10 + 3x^2 - 6x + 3)
2/15 √(x-1) (3x^2 + 4x + 8)
jiohoguygyx5arsfchyfy6rgvgdtye4wdyddyrdsrrsrsjydrjsrwaeqwa
To integrate the expression, x^2 / √(x - 1), we can use the substitution method. Let's start by assigning a new variable to the expression inside the square root.
Let u = x - 1
Differentiating both sides of the equation with respect to x, we get du/dx = 1.
Rearranging, we have dx = du.
Now, substitute these values into the integral:
∫ (x^2 / √(x - 1)) dx = ∫ ((u + 1)^2 / √u) du
Expanding the numerator, we have:
∫ (u^2 + 2u + 1) / √u du
Using the linearity of integration, we can split this into three separate integrals:
∫ (u^2 / √u) du + ∫ (2u / √u) du + ∫ (1 / √u) du
Simplifying each integral separately:
1. ∫ (u^2 / √u) du:
Rewrite the expression as u^(5/2) and then apply the power rule:
∫ u^(5/2) du = (2/7)u^(7/2) = (2/7)(x - 1)^(7/2)
2. ∫ (2u / √u) du:
Rewrite the expression as 2u^(3/2) and then apply the power rule:
∫ 2u^(3/2) du = (4/5)u^(5/2) = (4/5)(x - 1)^(5/2)
3. ∫ (1 / √u) du:
Rewrite the expression as u^(-1/2) and then apply the power rule:
∫ u^(-1/2) du = 2u^(1/2) = 2√(x - 1)
Summing up these integrals, we have:
∫ (x^2 / √(x - 1)) dx = (2/7)(x - 1)^(7/2) + (4/5)(x - 1)^(5/2) + 2√(x - 1) + C
Where C is the constant of integration.