What a nice integral!

Deriving the function $f(x)=x^{x^a}=e^{(\ln{x})x^a}$ , you get

$f′(x) = e^{(\ln{x})x^a}(\frac{1}{x}x^a +(\ln{x})ax^{a-1})=x^{x^a}x^{a-1}(1+a\ln{x})=x^{(x^a+a-1)}(1+a\ln{x})$

You can now see that the function $f′(x)$ is under the integral, so we can use calculus fundamental theorem to say

${\displaystyle \int_{1}^{2}x^{(x^{a}+a\text{-}1)}(1+a\ln{x})=\left.x^{x^{a}}\right|_{1}^{2}=2^{2^{a}}-1}$

Since the function $y=2^{2^x}$ is an increasing function, the equation $2^{2^{a}}-1=15$ has only the solution $a=2$ .

$\int_{1}^{2}{(1 +aln(x)){x}^{{x}^{a} + a - 1}}$

Substitute ${x}^{a} = u$

$\frac{1}{a}*\int_{1}^{{2}^{a}}{(1 + ln(u)){u}^{\frac{u}{a}}}$

Well, here I used the fact that $\frac{d}{dx}{x}^{\frac{x}{a}} = {x}^{\frac{x}{a}}\frac{(ln(x) + 1)}{a}$ . Well, everyone might want to consider that because we know on being 'derivated', it will give back itself "plus" "something".

Io, we have a very nice integral answer -

${u}^{\frac{u}{a}}$ , from $1$ to ${2}^{a}$ .

${2}^{{2}^{a}} - 1 = 15$

Hence, $a = 2$