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.]
