To use the Method of Partial Fractions, we let, for, #A,B,C in RR, : #
#x/((x-1)(x^2+4))=A/(x-1)+(Bx+C)/(x^2+4)............(1)#.
#A# can easily be obtained by using Heavyside's Cover-up Method :
#A=[x/(x^2+4)]_(x=1)=1/5#. Hence, with this #A#, from (1), we have,
#:. x/((x-1)(x^2+4))-(1/5)/(x-1)=(Bx+C)/(x^2+4)#, i.e.,
# (x-1/5(x^2+4))/((x-1)(x^2+4))=(Bx+C)/(x^2+4)#
#:. (cancel((x-1))(-1/5*x+4/5))/(cancel((x-1))(x^2+4))=(Bx+C)/(x^2+4)#.
#rArr B=-1/5, C=4/5#. Hence, by #(1)#, we have,
#intx/((x-1)(x^2+4))dx=1/5int1/(x-1)dx+int((-1/5*x+4/5))/(x^2+4)dx#
#=1/5ln|x-1|-1/10int(2x)/(x^2+4)dx+4/5int1/(x^2+4)dx#
#=1/5ln|x-1|-1/10int(d(x^2+4))/(x^2+4)+4/5*(1/2)arc tan(x/2)#
#=1/5ln|x-1|-1/10ln(x^2+4)+2/5arc tan (x/2)+K#.
Enjoy Maths.!
Note : The Method used to find #B, and, C# above, the following page from Wikipedia was used :
http://www.math-cs.gordon.edu/courses/ma225/handouts/heavyside.pdf
