Definite integral with a strange power I've been struggling for a while now on evaluating this disgusting integral:
$$(\ln2)\int_0^{(\ln2)^{1/\ln2}}2^{\ln x}\cdot x^\left(\frac{x^{\ln2}+1}{\ln x}-1\right)dx$$
My maths teacher gave our class this question a while ago, and he said that we should be able to do it (I am in high-school, and we have only been taught a fairly basic level of integration).
So today I spent many hours applying every integration technique I know to this monster, but I got absolutely nowhere. It got to a point where I couldn't think of another variable to use as a substitution because I had already made so many.  I eventually decided to plug this into an integral calculator and received a surprisingly nice result of $e$, however there was no further information and so I was not able to view any of the steps in how they got there. I am so stuck on this problem :(
Does anyone know how they got there? What are the steps in finding its indefinite form?
 A: we start by a well known notion $a=e^{\ln{(a)}}$
then
$$\frac{x^{\ln{(2)}}+1}{\ln x}-1$$
$$=\frac{x^{\ln{(2)}}+1-\ln{(x)}}{\ln{(x)}}$$
Now
$$x^{\frac{x^{\ln{(2)}}+1}{\ln x}-1}$$
$$=\left(e^{\ln{(x)}}\right)^\frac{x^{\ln{(2)}}+1-\ln{(x)}}{\ln{(x)}}$$
$$=e^{[x^{\ln{(2)}}+1-\ln{(x)}]}$$
$$=\frac{e^{[x^{\ln{(2)}}+1]}}{e^{\ln{(x)}}}$$
$$=\frac{e^{[x^{\ln{(2)}}+1]}}{x}$$
and since $2^{\ln{(x)}}=\left(e^{\ln{(2)}}\right)^{\ln{(x)}}=\left(e^{\ln{(x)}}\right)^{\ln{(2)}}=x^{\ln{(2)}}$ we get
$$2^{\ln{(x)}}\frac{x^{\ln{(2)}}+1}{\ln x}-1$$
$$=x^{\ln{(2)}}\frac{e^{[x^{\ln{(2)}}+1]}}{x}$$
$$=x^{[\ln{(2)}-1]}e^{[x^{\ln{(2)}}+1]}$$
From here on taking
$$u=x^{\ln{(2)}}+1$$
$$\Rightarrow du=\ln{(2)}x^{[\ln{(2)}-1]}dx$$
should change your integral into an $e^u$ integral.
A: This falls to some well-motivated $u$-substitutions:
\begin{align*}
\ln2\int 2^{\ln x} x^{\left(\frac{x^{\ln2}+1}{\ln x}-1\right)} dx
&= \ln 2 \int 2^t e^{2^t+1} dt 
    & \textrm{let $x=e^t$} \\
&= \int e^{s+1} ds 
    & \textrm{let $t=\log_2 s$}.
\end{align*}
In total this amounts to the $u$-substitution $x = s^{\log_2 e}$, which is bit more difficult to see.
A: $\newcommand{\d}{\mathrm{d}}$This is an extreme example of having more bark than bite. This is also a highly artificial integral, and everything just ‘magically’ works out.
You must convert everything to exp-log form. Begin: $$\begin{align}I&:=(\ln2)\int_0^{(\ln2)^{1/\ln2}}2^{\ln x}x^{\left(\frac{x^{\ln2}+1}{\ln x}-1\right)}\,\d x\\&=(\ln2)\int_0^{\exp(\ln\ln2/\ln2)}\exp\left((\ln x)(\ln2)+e^{(\ln2)(\ln x)}+1-\ln x\right)\,\d x\\&=\int_{-\infty}^{\ln\ln2}\exp(t+e^t+1)\,\d t\end{align}$$
Note $x^{a/\ln x}=\exp((\ln x)(a/\ln x))=\exp(a)$. I also substituted $t=(\ln2)(\ln x)$, you can check the details. That’s a natural substitution to make here.
Finally, let $\omega=e^t$, so that $\d\omega=e^t\,\d t$ and the RHS is present in our integrand. So we get: $$\int_0^{\ln2}\exp(\omega+1)\,\d\omega$$Which is just $e$.
A: Denote
$$f(x)=x^{\frac{x^{\ln 2}+1}{\ln x}}, x>0.$$
Let's calculate the derivative of $f$. Using the standard method, we differentiate the both sides of
$$\ln f(x)=\frac{x^{\ln 2}+1}{\ln x}\ln x=x^{\ln 2}+1,\tag{*}$$
to get
$$\frac{f'(x)}{f(x)}=\ln 2\cdot x^{\ln 2-1},$$
so
$$f'(x)=\ln 2\cdot x^{\ln 2-1}f(x).$$
We need to compute the integral
$$I=\int_0^{(\ln2)^{1/\ln2}}\ln 2\cdot2^{\ln x}x^{-1}f(x)\,dx.$$
Indeed, we have $2^{\ln x}=e^{\ln2\cdot \ln x}= x^{\ln 2}$, so the integral
$$I=\int_0^{(\ln2)^{1/\ln2}}\ln 2\cdot x^{\ln 2-1}f(x)\,dx=\int_0^{(\ln2)^{1/\ln2}}f'(x)\,dx=f\left((\ln2)^{1/\ln2}\right)-f(0+).$$
Now, we can easily get $f\left((\ln2)^{1/\ln2}\right)=2e$ and $f(0+)=e$ from $(*)$.
A: Rewrite the integrand as $\exp(f(\ln x))$ and simplify:
$$(\ln2)\int_0^{(\ln2)^{1/\ln2}}\exp(e^{\ln 2\ln x}+1-\ln x+\ln2\ln x)\,dx$$
$$=e\int_0^{(\ln2)^{1/\ln2}}\exp(e^{\ln 2\ln x})e^{\ln2\ln x}(\ln2)\frac1x\,dx$$
We see the factors produced by applying the chain rule when differentiating $\exp(e^{\ln2\ln x})$. The integrand has an elementary derivative:
$$=e[\exp(e^{\ln2\ln x})]_0^{(\ln2)^{1/\ln2}}=e(2-1)=e$$
