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integration- radicals

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Other answers:

yes but when i do that it still seems ugly
do one half at a time
i shud make a u sub \[u=1-x^7\]
in the 1st one
i'd say so
\[u=1-x^7\] \[\int\frac{u^{1/3}}{7x^6}du\]
the x^6 is a problem
maybe that method isn't going to work ugly as well but i think we are supposed to use 'by parts'
where did you get this question?
steward calculus advanved problem
what if we use the property integral ( f(x) from x=a to b = f(a+b-x) from x= a to b )?
in this case a+b=0+1 int f(1-x)=f(x)
yep, am not too sure if that'll work though, but probably should work ?
ok i have to go i;ll try it thanks to everyone for the effort
According to WA, the definite integral is 0: Maybe there's some detail we're not getting about the function itself.
interesting result maybe this property is because of the function is actually an inverse of itself i mean \[f(x)=\sqrt[3]{1-x^7}\] then \[f^{-1}(x)=\sqrt[7]{1-x^3}\] should this mean anything
\[\int_0^1 f(x)-f^{-1}(x) \space dx\] \[f(1)=f^{-1}(1)=0\]
this is a famous question on M.SE

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