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Hi, I've been trying to solve this integral for a while now, but I am still making no headway, any help would be much appreciated, thanks in advance.
[tex]\int\frac{1}{1-x^3}dx[/tex]
[tex]\int\frac{1}{1-x^3}dx[/tex]
Because it is incorrect.Chrono said:Couldn't you rewrite it as
[tex]\int {(1-x)^{-3}} dx[/tex]
and find the anti-dervative that way?
arildno said:Because it is incorrect.
Corrected:JonF said:Im not sure how you got those equalities, I get…
[tex]\frac{1}{1-x^3} = \frac{x + 2}{-x^4 -2x^2+x+2}[/tex]
JonF said:im not sure if this is right, but this is what i did...
[tex]\int\frac{1}{1-x^3}dx[/tex]
[tex]\int\frac{1}{(1-x)(1+x+x^2)}[/tex]
[tex]\int((1-x)(1+x+x^2))^{-1}[/tex]
you accidently got a negative on that two; otherwise you get the same as above.HallsofIvy said:x= -1 1= A+ (-B+C)(-2)= 1/3+ 2B- 4/3 so 2B= 2 or B= 1
lol, sorry I could easily solve this if you want but I was just trying to help the person who made the thread to be put on the right track.Parth Dave said:after looking at this again, if you complete the square on the denominator you obtain:
(x + 2)/[(x + 1/2)^2 + 3/4]
using a trig substitution you should be able to solve it
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