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check this: http://www.maths.usyd.edu.au/u/UG/JM/MATH1605/r/assign/standard_integrals.pdfdrynxz said:hmm ..theres use to be a standard intergral for this..but i dont think they teach it anymore.
i think its
int. dx/(a^2 - x^2) = 1/2a ln [(a+x)(a-x)] + c
and
int. dx/(x^2-a^2) = 1/2a ln [(x-a)/(x+a)] + c
int. dx/(a^2 - x^2) = 1/2a ln [(a+x)/(a-x)] + cdrynxz said:hmm ..theres use to be a standard intergral for this..but i dont think they teach it anymore.
i think its
int. dx/(a^2 - x^2) = 1/2a ln [(a+x)(a-x)] + c
and
int. dx/(x^2-a^2) = 1/2a ln [(x-a)/(x+a)] + c
You can always use partial fractions, as a last resort.TheCanuck said:Guys...how do u do these questions...Im sort of stuck here....
intergral of dx/94-x^2)
intergral of dx/(x^2-4)
thanx guys =)
woops its fixed nownickthebiscuit said:int. dx/(a^2 - x^2) = 1/2a ln [(a+x)/(a-x)] + c
yeah you missed the divide sign![]()