integrate x^3-2x^3+3x+2/x^5+2x^3+x dx?

To integrate the given expression, we can use partial fraction decomposition.

First, let's factor the denominator:
x^5 + 2x^3 + x = x(x^4 + 2x^2 + 1) = x(x^2 + 1)^2

Now, we can write the given expression as:
(x^3 - 2x^3 + 3x + 2) / (x(x^2 + 1)^2)

Next, we can perform partial fraction decomposition:
(x^3 - 2x^3 + 3x + 2) / (x(x^2 + 1)^2) = A/x + (Bx + C)/(x^2 + 1) + (Dx + E)/(x^2 + 1)^2

To find the values of A, B, C, D, and E, we can multiply both sides of the equation by the denominator:
(x^3 - 2x^3 + 3x + 2) = A(x^2 + 1)^2 + (Bx + C)x(x^2 + 1) + (Dx + E)x

Expanding and collecting like terms, we get:
x^3 - 2x^3 + 3x + 2 = A(x^4 + 2x^2 + 1) + (Bx^3 + Cx^2 + Bx + C) + (Dx^2 + Ex)

Comparing the coefficients of like terms on both sides, we have:
1 = A
0 = B + D
0 = C + B
0 = C + E
3 = B
2 = C

Therefore, A = 1, B = 3, C = 2, D = -3, and E = -2.

Substituting these values back into the partial fraction decomposition, we have:
(x^3 - 2x^3 + 3x + 2) / (x(x^2 + 1)^2) = 1/x + (3x + 2)/(x^2 + 1) - (3x + 2)/(x^2 + 1)^2

Now, we can integrate each term separately:
∫(x^3 - 2x^3 + 3x + 2) / (x(x^2 + 1)^2) dx = ∫(1/x) dx + ∫((3x + 2)/(x^2 + 1)) dx - ∫((3x + 2)/(x^2 + 1)^2) dx

Integrating each term, we get:
= ln|x| + 3/2 ln(x^2 + 1) - 3/(x^2 + 1) + C

Therefore, the integral of the given expression is ln|x| + 3/2 ln(x^2 + 1) - 3/(x^2 + 1) + C, where C is the constant of integration.