a) To find the values of \( A \), \( B \), and \( C \):
Given: \( 2x^3 – 3x + 1 = Ax \left( x^2 + 1 \right) + Bx + C \).
Expand the right-hand side:
\( Ax \left( x^2 + 1 \right) + Bx + C = Ax^3 + Ax + Bx + C \)
\( = Ax^3 + (A + B)x + C \)

Equate to the left-hand side: \( 2x^3 – 3x + 1 \):
\( Ax^3 + (A + B)x + C = 2x^3 – 3x + 1 \)

Compare coefficients:
– Coefficient of \( x^3 \): \( A = 2 \)
– Coefficient of \( x \): \( A + B = -3 \)
– Constant term: \( C = 1 \)

Solve for \( B \):
\( A + B = -3 \)
\( 2 + B = -3 \)
\( B = -5 \)

Thus:
\( A = 2 \), \( B = -5 \), \( C = 1 \) [2]

b) To find \( \int \frac{2x^3 – 3x + 1}{x^2 + 1} \, dx \):
From part (a), \( 2x^3 – 3x + 1 = 2x \left( x^2 + 1 \right) – 5x + 1 \).
Rewrite the integrand:
\( \frac{2x^3 – 3x + 1}{x^2 + 1} = \frac{2x \left( x^2 + 1 \right) – 5x + 1}{x^2 + 1} \)
\( = 2x – \frac{5x}{x^2 + 1} + \frac{1}{x^2 + 1} \)

Integrate term by term:
\( \int \left( 2x – \frac{5x}{x^2 + 1} + \frac{1}{x^2 + 1} \right) dx = \int 2x \, dx – \int \frac{5x}{x^2 + 1} \, dx + \int \frac{1}{x^2 + 1} \, dx \)

First term:
\( \int 2x \, dx = x^2 \)

Second term:
Let \( u = x^2 + 1 \), so \( du = 2x \, dx \), \( x \, dx = \frac{du}{2} \).
\( \int \frac{5x}{x^2 + 1} \, dx = \int \frac{5}{2u} \, du = \frac{5}{2} \ln u = \frac{5}{2} \ln (x^2 + 1) \)

Third term:
\( \int \frac{1}{x^2 + 1} \, dx = \arctan x \)

Combine:
\( \int \left( 2x – \frac{5x}{x^2 + 1} + \frac{1}{x^2 + 1} \right) dx = x^2 – \frac{5}{2} \ln (x^2 + 1) + \arctan x + c \)

Thus:
\( \int \frac{2x^3 – 3x + 1}{x^2 + 1} \, dx = x^2 – \frac{5}{2} \ln (x^2 + 1) + \arctan x + c \) [5]