<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] Put $x= \tan^{2}\theta$, then you have the integral as 
\begin{align*}
\int_{0}^{1} \sqrt{1+\frac{1}{x}} \ dx &= \int_{0}^{\pi/4} \sqrt{\frac{1+\tan^{2}\theta}{\tan^{2}\theta}} \cdot 2\tan\theta \cdot\sec^{2}\theta \ d\theta  \\\ &= 2 \cdot\int_{0}^{\pi/4} \sec^{3}\theta \ d\theta 
\end{align*}
Now integrate this function by parts. Take $u = \sec\theta$ then $du = \sec\theta \cdot \tan\theta$ and $dv = \sec^{2}\theta$. Then you have $v = \tan\theta$, so 
\begin{align*}
\int_{0}^{\pi/4} \sec^{3}\theta \ d\theta &= (\sec\theta\cdot\tan\theta)\:\biggl|_{0}^{\pi/4} - \int_{0}^{\pi/4} \sec\theta \cdot \tan^{2}\theta \ d\theta \\\ &= \frac{1}{\sqrt{2}} -\int_{0}^{\pi/4} \sec^{3}\theta \ d\theta + \int_{0}^{\pi/4} \sec\theta \ d\theta \\\ &= \frac{1}{2} \cdot \biggl\{ \frac{1}{\sqrt{2}} + \int_{0}^{\pi/4} \sec\theta \ d\theta \:\biggr\} \\\ &= \frac{1}{2\sqrt{2}} + \frac{1}{2} \cdot \bigl(\:\log(\sec\theta +\tan\theta)\bigr)_{0}^{\pi/4} \\\ &= \frac{1}{2\sqrt{2}} + \frac{1}{2} \cdot \log(\sqrt{2}+1)
\end{align*}
</s>