Let's perform the decomposition into partial fractions
#(x-2)/((x-1)(x-3)(x^2+5))=A/(x-1)+B/(x-3)+(Cx+D)/(x^2+5)#
#=(A(x-3)(x^2+5)+B(x-1)(x^2+5)+(Cx+D)(x-1)(x-3))/((x-1)(x-3)(x^2+5))#
Therefore,
#x-2=A(x-3)(x^2+5)+B(x-1)(x^2+5)+(Cx+D)(x-1)(x-3)#
Let #x=1#, #=>#, #-1=-12A#, #=>#, #A=1/12#
Let #x=3#, #=>#, #1=28B#, #=>#, #B=1/28#
Coefficients of #x^3#, #=>#, #0=A+B+C#
#C=-5/42#
Let #x=0#, #=>#, #-2=-15A-5B+3D#
#3D=15A+5B-2=15/12+5/28-2=-4/7#
#D=-4/21#
Therefore,
#(x-2)/((x-1)(x-3)(x^2+5))=(1/12)/(x-1)+(1/28)/(x-3)+(-5/42x-4/21)/(x^2+5)#
#=(1/12)/(x-1)+(1/28)/(x-3)-1/42(5x+8)/(x^2+5)#
So,
#I=int((x-2)dx)/((x-1)(x-3)(x^2+5))=1/12int(dx)/(x-1)+1/28int(dx)/(x-3)-1/42int((5x+8)dx)/(x^2+5)#
#=1/12ln(∣x-1∣)+1/28ln(∣x-3∣)-1/42(5int(xdx)/(x^2+5)+8int(dx)/(x^2+5))#
Let #u=x^2+5#, #=>#, #du=2xdx#
#int(xdx)/(x^2+5)=1/2 int (du)/u=1/2lnu=1/2ln(∣x^2+5∣)#
Let #v=x/sqrt5#, #=>#, #dv=dx/sqrt5#
#8int(dx)/(x^2+5)=8/sqrt5int(dv)/(v^2+1)#
Let #v=tantheta#, #dv=sec^2theta d theta#
and #1+tan^2theta=sec^2theta#
So,
#8/sqrt5int(dv)/(v^2+1)=8/sqrt5int(sec^2theta/sec^2thetad theta)#
#=8/sqrt5theta=8/sqrt5arctanv=8/sqrt5arctan(x/sqrt5)#
Putting it all together
#I=1/12ln(∣x-1∣)+1/28ln(∣x-3∣)-5/84ln(x^2+5)- 4/(21sqrt5)arctan(x/sqrt5)+C#
