First we should get in the numerator a polynomial of a grade inferior than the denominator's
By long division:
" "2x^3-3x^2-4x-3" "|" "-6x^2+3x-4
-2x^3+x^2-4/3x" "|____
_____"
"-1/3x+1/3
" "-2x^2-16/3x-3
" "2x^2-x+4/3
" " _______
" "-19/3x-5/3=-1/3(19x+5)
So the expression becomes
=1/3int (-x+1)dx-1/3*1/6int(19x+5)/(x^2-x/2+2/3)dx
Dealing with
=int(19x+5)/(x^2-x/2+2/3)dx
Since (x-1/4)^2=x^2-x/2+1/16 and 2/3-1/16=29/48=(1/4sqrt(29/3))^2
We can make
x-1/4=1/4sqrt(29/3)*tan y
=> dx=1/4sqrt(29/3)*sec^2 y*dy
Now let's find how many units of (x-1/4) there are in the denominator
" "19x+5" "|" "x-1/4
-19x+19/4" "|____
_____"
"19
" "39/4
So the partial expression becomes
=19int ((1/4sqrt(29/3)tan y)(1/4sec^2 y))/(1/16*29/3sec^2 y)dy+39/4int (1/4sqrt(29/3))/(1/16*29/3sec^2 y)dy
=-19ln|cos y|+39sqrt(3/29)y
But tan y = sqrt(3/29)(4x-1) => siny =sqrt(3/29)(4x-1)cosy
And sin^2y+cos^2y=1 => (3/29(16x^2-8x+1)+1)cos^2 y=1 => (48x^2-24x+32)/29*cos^2 y=1 => cos y=(1/2sqrt(29/2))/sqrt(6x^2-4x+8
Then the partial expression becomes
=-19ln|(1/2sqrt(29/2))/sqrt(6x^2-4x+8)+39sqrt(3/29)tan^(-1) (sqrt(3/29)(4x-1))
Back to the main expression:
=-x^2/6+1/3-1/18[-19ln|(1/2sqrt(29/2))/sqrt(6x^2-4x+8)+39sqrt(3/29)tan^(-1) (sqrt(3/29)(4x-1))]
=-x^2/6+x/3+19/18ln|(1/2sqrt(29/2))/sqrt(6x^2-3x+4)|-13/(2sqrt(87))*tan^(-1) (4sqrt(3/29)(x-1/4))+const.
One step further: simplifying
The result above is good enough, but the constant part of the logarithm can be absorbed by the general constant:
=-x^2/6+x/3+19/18[ln(1/2sqrt(29/2))-(1/2)ln|6x^2-3x+4|]-13/(2sqrt(87))*tan^(-1) (4sqrt(3/29)(x-1/4))+const.
=-x^2/6+x/3-19/36ln|6x^2-3x+4|-13/(2sqrt(87))*tan^(-1) (4sqrt(3/29)(x-1/4))+const.