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?
 A: A good thing to note here is that $y = 2 \sqrt x$ is the same as $x = \frac{y^2}{4}$. So by 'swapping the order of integration' (sort of), you calculate a much easier integral. But do remember that the domain for $x$ here is $[0,1]$, but for $y$ it's $[0,2]$.

A: $$u=1+\frac{1}{x}\Longrightarrow x\to 0^+\longrightarrow u\to +\infty\,\,,\,\,x=1\longrightarrow u = 2$$ so with the new integration limits we get $$\int_\infty^2-\sqrt{u}\,\left(-\frac{du}{(u-1)^2}\right)=\int_2^\infty\frac{\sqrt{u}}{(u-1)^2}\,du$$
A: 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*}
A: Here's something you might try. Note that the length of that arc will be the same as the length of the same arc, reflected over the line $y=x$. That is, the arc $y=x^2/4$, from $x=0$ to $x=2$.
A: Substitute $u=\sqrt{1+\frac{1}{x}}=\sqrt{\frac{x+1}{x}}$. Then $x=\frac{1}{u^2-1}$, so $dx=-\frac{2u}{(u^2-1)^2}du$, which makes: 
$$\int\sqrt{1+\frac{1}{x}}\,{dx}=-\int \frac{2u^2}{(u^2-1)^2}\,{du}$$
Can you continue from here?
A: You can try $x = \cot^2(\theta), dx = -2\cot(\theta) \csc^2(\theta)$. This substitution comes from knowing that $\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \dfrac{1}{\cot^2(\theta)}$. The final integral is quite simple depending upon your comfort with trigonometry.
A: 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
$$\sqrt{2} + {1 \over 2}\ln\bigg({y - 1 \over y + 1}\bigg)\bigg|_{\sqrt{2}}^{\infty}$$
$$=\sqrt{2} - {1 \over 2}\ln\bigg({\sqrt{2} - 1 \over \sqrt{2} + 1}\bigg)$$
A: 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}
A: \begin{align*}
\int_0^1 \sqrt{1+\frac{1}{x}}\,{\rm d}x&=\int_0^1 \sqrt{1+\frac{1}{t^2}}\,{\rm d}t^2\\
&=2\int_0^1 \sqrt{t^2+1}\,{\rm d}t\\
&=2\int_0^{{\rm arsinh}\ 1} \sqrt{\sinh^2 u+1}\,{\rm d}(\sinh u)\\
&=2\int_0^{{\rm arsinh}\ 1} \cosh^2u\,{\rm d}u\\
&=\int_0^{{\rm arsinh}\ 1} \cosh 2u+1\,{\rm d}u\\
&=\left[\frac{1}{2}\sinh 2u+u\right]_0^{{\rm arsinh}\ 1}\\
&=\sqrt{2}+\ln(1+\sqrt2).
\end{align*}
A: Notice that the integrand is a differential binomial, and then you may apply "Integration of differential binomial" (P. L. Chebyshev) and you're immediately done. See here.
