<s>[INST] $\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx$ I was attempting to solve an MIT integration bee problem (1) when I misread the integral and wrote (2) instead. 
$$\int\sqrt{x\cdot \sqrt[3]{x\cdot \sqrt[4]{x\cdot\sqrt[5]{x\ldots } }}}\,dx\tag{1}$$
$$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx\tag{2}$$
I was able to solve (1), as the integrand simplifies to $x^{e-2}$, however, I'm struggling with solving (2). 
If we rewrite the roots as powers, we get:
$$\int x^\frac{2}{2}\cdot x^\frac{3}{4}\cdot x^\frac{4}{8}\cdot x^\frac{5}{16}\ldots\,dx$$
combining the powers we get:
$$\int x^{\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\ldots}$$
the exponent is the infinite sum
$$\sum^{\infty}_{n=1}\frac{n+1}{2^n}\tag{3} $$
we can split this into: 
$$\sum^{\infty}_{n=1}\frac{n}{2^n}+\sum^{\infty}_{n=1}\frac{1}{2^n} $$
The right sum is well known except here the sum begins at $n=1$, meaning that the right sum evaluates to 1. Messing around with desmos, the integrand appears to be $x^3,x>0$ implying that (3) converges to 3 and the $\sum^{\infty}_{n=1}\frac{n}{2^n}$ converges to 2.
Which is part I'm struggling with. Any ideas? 
$$\sum^{\infty}_{n=1}\frac{n}{2^n}$$
 [/INST] The binomial theoem gives$$(1-x)^{-2}=\sum_{n\ge0}(n+1)(-1)^nx^n,$$so the exponent is$$-1+(1-1/2)^{-2}=3.$$So the integral is $\int x^3dx=\frac14x^4+C$.
</s>