We first factorise the #Dr.#
#Dr.=x^3-5x^2+x-5= x^2(x-5)+1(x-5)=(x-5)(x^2+1).#
Thus, #Dr.# has a linear factor = (x-5), and a quadratic one = (x^2+1), which can not be further factorised [ in #RR#]. In such a case, we need to find #A,B,C in RR# under the following supposition ::
#(10x+2)/(x^3-5x^2+x-5)=(10x+2)/{(x-5)(x^2+1)}=A/(x-5)+(2Bx+C)/(x^2+1).................(1)#
The #2# in #(2Bx+C)# has been so chosen, bcz. #d/dx(x^2+1)=2x.#
Simplifying the #R.H.S.# of #(1)# and then comparing the #Nrs.# of both sides, we find that,
#A(x^2+1)+(2Bx+C)(x-5)=10x+2...............(2)#
To determine #A,B,C# of #(2)#, we can take any #three# values of x; sub. them one by one in #(2)# giving #3# eqns. in #A,B,C# and solve them.
But, to make computational work easier, we tactfully choose #x=5,0,1,# to see :
#x=5 rArr 26A=52 rArr A=2.#
#x=0 rArr A-5C=2 rArr 2-5C=2 rArr C=0.#
#x=1rArr2A+(2B+C)(-4)=12rArr4-8B=12rArr B=-1.#
Using these #A,B,C# in #(1),# we have,
#int(10x+2)/(x^3-5x^2+x-5)dx=int[2/(x-5)-(2x)/(x^2+1)]dx=int2/(x-5)dx-int(d/dx(x^2+1))/(x^2+1)dx=2ln|x-5|-ln(x^2+1)+K.#
[#K# const. of int.]
In the second integral, we have made use of the formula : #int(f'(x))/f(x)dx=ln|f(x)|+C_1.#[#C-1# const. of int.]
