Let I=int(x^2-3x)/{(x-1)(x+2)}dx=int(x^2-3x)/(x^2+x-2)dx.I=∫x2−3x(x−1)(x+2)dx=∫x2−3xx2+x−2dx.
The Degree of Poly. in Nr.Nr. = 22 = that of poly in Dr.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+2Nr.=x2−3x=x2+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)x2−3xx2+x−2=(x2+x−2)−(4x−2)x2+x−2
=(x^2+x-2)/(x^2+x-2)-(4x-2)/(x^2+x-2)=1-(4x-2)/(x^2+x-2)=x2+x−2x2+x−2−4x−2x2+x−2=1−4x−2x2+x−2
=1-(4x-2)/{(x-1)(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,I=∫[1−4x−2(x−1)(x+2)]dx=∫1dx−I1=x−I1,
where I_1=int(4x-2)/{(x-1)(x+2)}dxI1=∫4x−2(x−1)(x+2)dx
To evaluate I_1I1, 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
