How do you integrate #10/[(x-1)(x^2+9)] dx#?

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Answer 1 · 2015-06-17T13:31:02

#10/((x-1)(x^2+9)) = 1/(x-1) + (-x-1)/(x^2+9) = 1/(x-1) - x/(x^2+9) - 1/(x^2+9)#

#10/((x-1)(x^2+9)) = A/(x-1) + (Bx+C)/(x^2+9)#

So we want: #Ax^2+9A +Bx^2-Bx+Cx-C = 10#
Hence:
#A+B =0#
#-B+C=0#
#9A-C=10#

The second equation gives us: #B=C#, so
#A+C =0#
#9A-C=10#, adding gets us:

#10A=10#, so #A=1# and #B=C =-1#

And #10/((x-1)(x^2+9)) = 1/(x-1) + (-x-1)/(x^2+9) = 1/(x-1) - x/(x^2+9) - 1/(x^2+9)#,

so

#int 10/((x-1)(x^2+9)) dx = int 1/(x-1) dx - int x/(x^2+9) dx - int1/(x^2+9) dx#

#= ln abs(x-1) -1/2 ln(x^2+9) - 1/3 tan^(-1) (x/3) +C#