First, separate it into the format:
#A/(x-5) + B/(x-3)#
Now cross-multiply and simplify:
#(A(x-3) + B(x-5))/((x-5)(x-3))#
#= (Ax - 3A + Bx - 5B)/((x-5)(x-3))#
#= ((A + B)x - 3A - 5B)/((x-5)(x-3))#
Notice there's an #x# term and an #x^0# term (constants). We can equate them back to what the numerator originally was. There was no #x# term originally, so:
#A + B = 0#
#A = -B#
#-3A - 5B = 2#
# => 3B - 5B = 2#
#-2B = 2#
# => B = -1#
# => A = 1#
You're basically done. I was taught that #int1/xdx = ln|x| + C#... so:
#= int 1/(x-5) - 1/(x-3)dx#
#= ln|x-5| - ln|x-3| + C#
Wolfram Alpha doesn't seem to agree with me about absolute values here compared to here, so let's just try differentiating this to see what we get.
#[(df(x))/(dx)] " at " x > 0 = 1/(x-5) - 1/(x-3)#
#= ((x-3) - (x-5))/((x-5)(x-3))#
#= (x-3 - x+5)/((x-5)(x-3))#
#= 2/((x-5)(x-3))#
...Yup, it works. #"*shrug*"#
