Hi. How can I integrate 1/(X^3 +1) ? Thank you to anyone who can help me :-)
Write 1/(x^3 +1) as
1/[(x+1)(x^2-x+1)]
Then use integration by parts, letting
dv = dx/(x^2 -x +1)
u = 1/(x+1)
du = log (x+1)
v = (2/sqrt3)arctan[(2x-1)/sqrt3]
That should take you to the answer. I know what the answer is (from my table of integrals), but have not done all the steps.
Could you by any chance tell me what your table of integrals says. I can't find it on mine to check my answer
OK. The answer is
(1/3){(1/2)log[(x+1)^2/(x^2-x+1)] + sqrt3*arctan[(2x-1)/sqrt3]}
Quite a mess. My approach does seem to lead to that answer.
To integrate 1/(x^3 + 1), you can follow these steps:
1. Rewrite the expression as 1/[(x+1)(x^2-x+1)].
2. Use integration by parts, which involves choosing two parts of the function to differentiate and integrate. Let's choose:
dv = dx/(x^2 - x + 1)
u = 1/(x+1)
3. Calculate the differentials:
du = dx/(x+1)