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∫[2^(X^(2))x,]dx=
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Try letting u=x^2
du=2xdx then.
or you can say du/2=xdx
If y=a^x, lny=ln(a^x)=xlna => y'/y=lna => y'=y(lna)=(a^x)lna Integrate both sides to get y=int((a^x)lna,x)=lna(int(a^x,x)) => y/lna=int(a^x,x) => (Remember y=a^x) a^x/lna=int(a^x,x) (Don't forget about your constant C) => int(a^x,x)=a^x/lna + C. So if you have 1/2*int(2^u,u)=1/2*(2^u/ln2)+C (Almost done! Put back in terms of x)
Hey that does say int(2^(x^(2)), x)? right?
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