How can you find the antiderivative of x^x^x.?

Question

I know it involves the natural log of something.

Answer 1

I am assuming that there ar eonly three x's because usually this problem is giving with an infinite number of x's looking like
y=x^x^x^x^x...

For this one y=x^x^x, all you do is take natural log of both sides and you get,
ln(y)=(x^x)*ln(x)

Now take the derivative of both sides and use the product rule on the right hand side.

y'/y=d(x^x)/dx * ln(x) + (x^x)/x

For the first term, do the same thing,
let z=x^x
ln(z)=x(ln(x))
z'/z=ln(x)+1

multiply both sides by z
z'=z*ln(x)+z

but z is just x^x
z'=(x^x)(ln(x)+1)

So now go back up and plug it in
y'/y=d(x^x)/dx * ln(x) + (x^x)/x

y'/y=(x^x)(ln(x)+1)*ln(x)+(x^x)/x

Multiply both sides by y which is just x^x^X
y'=((x^x)(ln(x)+1)*ln(x)+(x^x)/x)*x^x^x

And I am sure that this can be simplified.

Answer 2

Sorry, this antiderivative cannot be expressed
in terms of elementary functions. For that
matter, neither can the antiderivative of x^x

Answer 3

good luck with that