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?

Note that the integrand $\sqrt{1 + {1 \over x}}$ decreases from infinity to $\sqrt{2}$ as $x$ goes from $0$ to $1$. The area under the graph is therefore equal to the area of the box $[0,1] \times [0,\sqrt{2}]$ plus the area under the graph of the inverse function $g(y)$ to $\sqrt{1 + {1 \over x}}$ from $y = \sqrt{2}$ to $y = \infty$. Note that $g(y) = {1 \over y^2 - 1}$. So the answer is
$$\sqrt{2} + \int_{\sqrt{2}}^{\infty} {1 \over y^2 - 1}\,dy$$
This integral is easily computed, using partial fractions for example. The result is
$$\sqrt{2} + {1 \over 2}\ln\bigg({y - 1 \over y + 1}\bigg)\bigg|_{\sqrt{2}}^{\infty}$$
$$=\sqrt{2} - {1 \over 2}\ln\bigg({\sqrt{2} - 1 \over \sqrt{2} + 1}\bigg)$$
