Let #I=int(x^2-3x)/{(x-1)(x+2)}dx=int(x^2-3x)/(x^2+x-2)dx.#
The Degree of Poly. in #Nr.# = #2# = that of poly in #Dr.#
Hence, it is Improper Rational fun. To make it Proper, usually Long Division is performrd, but, here we proceed as under :
We write, #Nr.=x^2-3x=x^2+x-2 [i.e., = Dr]-4x+2#, so
#(x^2-3x)/(x^2+x-2)={(x^2+x-2)-(4x-2)}/(x^2+x-2)#
#=(x^2+x-2)/(x^2+x-2)-(4x-2)/(x^2+x-2)=1-(4x-2)/(x^2+x-2)#
#=1-(4x-2)/{(x-1)(x+2)}.#
Hence, #I=int[1-(4x-2)/{(x-1)(x+2)}]dx=int1dx-I_1=x-I_1,#
where #I_1=int(4x-2)/{(x-1)(x+2)}dx#
To evaluate #I_1#, we have to split #(4x-2)/{(x-1)(x+2)}......(1)# using Partial Fractions as #A/(x-1)+B/(x+2)={A(x+2)+B(x-1)}/{(x-1)(x+2)}.....(2); A,B in RR.#
Since #(1) and (2)# are equal, we get #: A(x+2)+B(x-1)=4x-2, AAx.#
In particular, #x=1 rArr 3A=2 rArr A=2/3# and #x=-2 rArr B=10/3.#
Accordingly, #I_1=int(2/3)/(x-1)dx+int(10/3)/(x+2)dx#
#=2/3ln|x-1|+10/3ln|x+2|.# Finally, we altogether get,
#I=x-2/3ln(x-1)-10/3ln(x+2)+C#, or, using the Rules of Log,
#I=x-2/3ln(x-1)(x+2)^5+C=x-ln(x-1)^(2/3)(x+2)^(10/3)+C#
