1 - 12/(x-2) + 17/(x-3)
You lost the 1.
Other than that, it's ok.
integrate (x^2+8)/(x^2-5x+6) by partial fractions method.
I first did long division, and came up with 1+((5x+2)/(x^2-5x+6)). Then I did partial fraction for (5x+2)/(x^2-5x+6) by going A/(x-3)+B/(x-2) and I got 17/(x-3)-12/(x-2) to take the integral of. But when I check online with wolfram, the answer is different, with what i got: x+ 17ln(x-3)-12ln(x-2)+c.
Please explain and help step by step. Thanks!
3 answers
I actually didn't lose the 1, if you look at my final answer.
Right. I misread you r text.
But, wolfram's answer looks the same to me, if you allow that
∫ du/u = ln|u|
But, wolfram's answer looks the same to me, if you allow that
∫ du/u = ln|u|