Let #f(x)=2x^3-5x^2+x+2#
#f(1)=2-5+1+2=0#
Therefore, #(x-1)# is a factor
#f(2)=16-20+2+2=0#
Therefore, #(x-2)# is a factor
#(x-1)(x-2)=x^2-3x+2# is a factor
To find the last factor, we do a long division
#color(white)(aaaa)##2x^3-5x^2+x+2##color(white)(aaaa)##∣##x^2-3x+2#
#color(white)(aaaa)##2x^3-6x^2+4x##color(white)(aaaaaaa)##∣##2x+1#
#color(white)(aaaaaa)##0+x^2-3x+2#
#color(white)(aaaaaaaa)##+x^2-3x+2#
#color(white)(aaaaaaaaaa)##+0-0+0#
We can perform the decomposition into partial fractions
#(x^2+1)/(2x^3-5x^2+x+2)=(x^2+1)/((x-1)(x-2)(2x+1))#
#=A/(x-1)+B/(x-2)+C/(2x+1)#
#=(A(x-2)(2x+1)+B(x-1)(2x+1)+C(x-1)(x-2))/((x-1)(x-2)(2x+1))#
Therefore,
#x^2+1=A(x-2)(2x+1)+B(x-1)(2x+1)+C(x-1)(x-2))#
Let #x=1#, #=>#, #2=-3A#, #=>#, #A=-2/3#
Let #x=2#, #=>#, #5=5B#, #=>#, #B=1#
Let #x=-1/2#, #=>#, #5/4=15/4C#, #=>#, #C=1/3#
So,
#(x^2+1)/(2x^3-5x^2+x+2)=(-2/3)/(x-1)+1/(x-2)+(1/3)/(2x+1)#
#int((x^2+1)dx)/(2x^3-5x^2+x+2)=int((-2/3)dx)/(x-1)+int(dx)/(x-2)+int((1/3)dx)/(2x+1)#
#=-2/3ln(∣x-1∣)+ln(∣x-2∣)+1/6ln(∣2x+1∣)+C#
