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