<s>[INST] How to evaluate $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$ I need to calculate the length of a curve $y=2\sqrt{x}$ from $x=0$ to $x=1$.
So I started by taking $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$, and then doing substitution: $\left[u = 1+\frac{1}{x}, \text{d}u = \frac{-1}{x^2}\text{d}x \Rightarrow -\text{d}u = \frac{1}{x^2}\text{d}x \right]^1_0 = -\int\limits^1_0 \sqrt{u} \,\text{d}u$ but this obviously will not lead to the correct answer, since $\frac{1}{x^2}$ isn't in the original formula. 
Wolfram Alpha is doing a lot of steps for this integration, but I don't think that many steps are needed.
How would I start with this integration?
 [/INST] You can use integrate by parts:
\begin{align}
I&=\int_0^1\sqrt{1+\frac{1}{x}}dx=x\sqrt{1+\frac{1}{x}}|_0^1-\int_0^1x\frac{-\frac{1}{x^2}dx}{2\sqrt {1+\frac{1}{x}}}
=\sqrt{x(1+x)}|_0^1+\int_0^1\frac{d\sqrt{x}}{\sqrt {1+x}}\\
&=\sqrt{2}+\ln(\sqrt{x}+\sqrt{x+1})|_0^1
=\sqrt{2}+\ln(1+\sqrt{2})
\end{align}
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