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

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

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

How do you integrate 10/[(x-1)(x^2+9)] dx10/[(x-1)(x^2+9)] dx?