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.#
