How do i integrate (x^3)/(x-1)??? please help me out...
2007-09-30 15:29:34 · 5 answers · asked by tumbleweedyg 2 in Science & Mathematics ➔ Mathematics
since the degree of the numerator (3) is larger than that of the denominator (1) we have to do long division.
It's hard to write out the work but you'll get x^2 + x + 1 with a remainder of -1 (ask if you need help with this part)
So our new formula to integrate is x^2 + x + 1 - (x-1)^-1
This should be much easier to integrate:
(x^3)/3 + (x^2)/2 + x - ln(x-1) + c
hope that helps
2007-09-30 15:42:09 · answer #1 · answered by Jaron 2 · 1⤊ 0⤋
This divides out to x²+x+1+1/(x-1)
integrate separately
x^3/3+x²/2+x+ln(x-1)+c
2007-09-30 22:38:15 · answer #2 · answered by chasrmck 6 · 0⤊ 0⤋
I did it by repeated parts and got
(1/6)(x-1)(x(2x + 5) + 11) + ln(x-1) + C
It's kinda sticky and takes a few minutes, but it works âº
Doug
2007-09-30 22:37:23 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋
x^3/(x-1)
(x^3 - 1 + 1)/(x-1)
(x^3-1)/(x-1) + 1/(x-1)
(x-1)(x^2 +x + 1)/(x - 1) + 1/(x-1)
x^2 + x + 1 + 1/(x-1)
â«x^3/(x-1) dx = â«x^2 dx+ â«x dx + â«dx + â«dx/(x-1)
x^3/3 + x^2/2 + x + ln(x-1) + c
2007-09-30 22:38:45 · answer #4 · answered by mohanrao d 7 · 0⤊ 0⤋
What kind of help do you need?
2007-09-30 22:32:13 · answer #5 · answered by Anonymous · 0⤊ 1⤋