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]