Calculus: Early Transcendentals, Chapter 6, 6.3, Section 6.3, Problem 24

The shell has the radius x - (-1) , the cricumference is 2pi*(x+1) and the height is (2x)/(1+x^3) - x, hence, the volume can be evaluated, using the method of cylindrical shells, such that:
V = 2pi*int_(x_1)^(x_2) (x+1)*((2x)/(1+x^3) - x) dx
You need to find the endpoints, using the equation (2x)/(1+x^3) - x = 0 => 2x - x - x^4 = 0 => 3x - x^4 = 0 => x(3 - x^3) = 0 => x = 0 and x = root(3) 3
V = 2pi*int_0^(root(3) 3) (x+1)*((3x - x^4)/(1+x^3)) dx
V = 2pi*int_0^(root(3) 3) (x+1)*(3x - x^4)/((x+1)(x^2-x+1)) dx
Reducing the like terms yields:
V = 2pi*int_0^(root(3) 3) (3x - x^4)/(x^2-x+1) dx
V = 2pi*(int_0^(root(3) 3) (3x)/(x^2-x+1) -int_0^(root(3) 3) (x^4)/(x^2-x+1) dx
V = 2pi*(int_0^(root(3) 3) (3x)/(x^2-x+1)dx - int_0^(root(3) 3) x^2 dx - int_0^(root(3) 3) x dx + int_0^(root(3) 3) (x)/(x^2-x+1)dx)
V = 2pi*(int_0^(root(3) 3) (4x)/(x^2-x+1)dx - int_0^(root(3) 3) x^2 dx - int_0^(root(3) 3) x dx)
V = 2pi*(2*int_0^(root(3) 3) (2x+1-1)/(x^2-x+1)dx - x^3/3|_0^(root(3) 3) - x^2/2|_0^(root(3) 3))
V = 2pi*(2*int_0^(root(3) 3) (2x-1)/(x^2-x+1)dx+ 2*int_0^(root(3) 3) (1)/(x^2-x+1)dx - 1 - (root(3) 9)/2)
You need to solve int_0^(root(3) 3) (2x-1)/(x^2-x+1)dx using substitution x^2-x+1 = t => (2x-1)dx =dt.
V = 2pi*(2*ln(x^2-x+1)|_0^(root(3) 3)+ 2*int_0^(root(3) 3) (1)/((x-1/2)^2 + ((sqrt3)/2)^2)dx - 1 - (root(3) 9)/2)
V = 2pi*(2*ln((root(3) 9)-(root(3) 3)+1)+ (4/sqrt3)*arctan (2x-1)/sqrt3|_0^(root(3) 3) - 1 - (root(3) 9)/2)
V = 2pi*(2*ln((root(3) 9)-(root(3) 3)+1)+ (4/sqrt3)*arctan (2(root(3) 3)-1)/sqrt3 -(4/sqrt3)*(pi/6) - 1 - (root(3) 9)/2)
Hence, evaluating the volume, using the method of cylindrical shells, yields V = 2pi*(2*ln((root(3) 9)-(root(3) 3)+1)+ (4/sqrt3)*arctan (2(root(3) 3)-1)/sqrt3 -(4/sqrt3)*(pi/6) - 1 - (root(3) 9)/2)

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