The decomposition in partial fractions is
#(3x^2+x+4)/((x^2+2)(x^2+1))=(Ax+B)/(x^2+2)+(Cx+D)/(x^2+1)#
#3x^2+x+4=(Ax+B)(x^2+1)+(Cx+D)(x^2+2)#
Let #x=0# then 4= B+2D
Coefficients of #x^2#, #3=B+D#
From thes equations. we get #D=1# and #B=2#
So,
#3x^2+x+4=(Ax+2)(x^2+1)+(Cx+1)(x^2+2)#
We compare the coefficients of #x#
#1=A+2C#
and coefficients of #X^3#, #0=A+C#
So, #A=-1# and #C=1#
#(3x^2+x+4)/((x^2+2)(x^2+1))=(-x+2)/(x^2+2)+(x+1)/(x^2+1)#
#int((3x^2+x+4)dx)/((x^2+2)(x^2+1))=int((-x+2)dx)/(x^2+2)+int((x+1)dx)/(x^2+1)#
#=int(-xdx)/(x^2+2)+int(2dx)/(x^2+2)+int(xdx)/(x^2+1)+intdx/(x^2+1)#
#=-ln(x^2+2)/2+ln(x^2+1)/2+arctanx+sqrt2arctan(x/sqrt2)+C#
as#intdx/(x^2+1)=arctanx#
and #int(xdx)/(x^2+1)=ln(x^2+1)/2#
