Calculating Mss of a Rod using Integration

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Homework Statement



Let a rod with length L and constant cross-sectional area A have the density
ρ(x) = ρ((3.x^2 + 2.x.L)/L^2)

0 ≤ x ≤ L

where x is the distance from one end of the rod and ρ0 is a real constant.
(a) Find the total mass M of the rod.
(b) Find the x-coordinate xc of the centre of mass of the rod.
(c) Find, in terms of M, the moment of inertia of the rod about the vertical axis through x = 0.


Homework Equations



M = ∫Aρ(x).dx

The Attempt at a Solution



M = ∫Aρ(x).dx

M= A.ρ∫((3.x^2 + 2.x.L)/L^2).dx

I am slightly confused where to go from here! Do I use integration by parts?
 
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Why? That is just the integral of a polynomial. What is the integral of 3x^2 dx? What is the integral of 2x dx?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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