Given that \(2{x^3} - 3x + 1\) can be expressed in the form \(Ax\left( {{x^2} + 1} \right) + Bx + C\), find the values of the constants \(A\), \(B\) and \(C\).

Hence find \(\int {\frac{{2{x^3} - 3x + 1}}{{{x^2} + 1}}} {\text{d}}x\).

\(2{x^3} - 3x + 1 = Ax\left( {{x^2} + 1} \right) + Bx + C\)

\(A = 2,\,\,C = 1,\)     A1

\(A + B =  - 3 \Rightarrow B =  - 5\)     A1

[2 marks]

\(\int {\frac{{2{x^3} - 3x + 1}}{{{x^2} + 1}}} {\text{d}}x = \int {\left( {2x - \frac{{5x}}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}} \right)} {\text{d}}x\)      M1M1

Note: Award M1 for dividing by \(\left( {{x^2} + 1} \right)\) to get \(2x\), M1 for separating the \(5x\) and 1.

\( = {x^2} - \frac{5}{2}{\text{ln}}\left( {{x^2} + 1} \right) + {\text{arctan}}\,x\left( { + c} \right)\)     (M1)A1A1

Note: Award (M1)A1 for integrating \({\frac{{5x}}{{{x^2} + 1}}}\), A1 for the other two terms.

[5 marks]