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\[\color{blue}{\huge\int_0^1\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}dx}\]

\[\color{blue}{\int_0^1\sqrt[3]{1-x^7}dx-\int\sqrt[7]{1-x^3}dx}\]

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

\[\int\frac{u^{1/3}}{7(1-u)^{1/7}}\]

hmm,

maybe that method isn't going to work

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

\[\int_0^1 f(x)-f^{-1}(x) \space dx\]
\[f(1)=f^{-1}(1)=0\]

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