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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?

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If $u=1+x$, then $du$ is simply $dx$. – Gigili May 28 '12 at 12:36
Correct - I made some TeX mistakes, and realized why the subsitution won't work 10 seconds after posting. Should've been fixed now – Thom Wiggers May 28 '12 at 12:39
Keep editing, you'll get it right eventually. – Gerry Myerson May 28 '12 at 12:39
Don't forget that you have to change the limits of integration when you do a substitution. – Gerry Myerson May 28 '12 at 12:40
Then it's not also $-\int\limits^1_0 \sqrt{u} \,\text{d}u$, Since $dx=-x^2 du$. – Gigili May 28 '12 at 12:44
up vote 11 down vote accepted

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 $x=2$.

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Surely the easiest way - one might also argue $y=2\sqrt{x}$ gives $x=y^2/4$. – AD. May 28 '12 at 15:57
@AD.: True enough. It amounts to the same thing, of course. – Cameron Buie May 28 '12 at 16:31
Yes, I just thought it should be emphasised. :) – AD. May 28 '12 at 20:10
It fees like a dirty hack. I love it :P – Thom Wiggers May 29 '12 at 8:00
@TheGuyOfDoom: Glad you approve! – Cameron Buie May 31 '12 at 16:30

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?

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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 $[0,2]$.

enter image description here

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You mean $x=\cfrac{y^2}{4}$. – Cameron Buie May 28 '12 at 14:44
@Cameron: Thanks for that - – mixedmath May 28 '12 at 15:35

$$u=1+\frac{1}{x}\Longrightarrow x\to 0^+\longrightarrow u\to +\infty\,\,,\,\,x=1\longrightarrow u = 2$$ so with the new integration limits we get $$\int_\infty^2-\sqrt{u}\,\left(-\frac{du}{(u-1)^2}\right)=\int_2^\infty\frac{\sqrt{u}}{(u-1)^2}\,du$$

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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*}

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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.

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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 $$\sqrt{2} + {1 \over 2}\ln\bigg({y - 1 \over y + 1}\bigg)\bigg|_{\sqrt{2}}^{\infty}$$ $$=\sqrt{2} - {1 \over 2}\ln\bigg({\sqrt{2} - 1 \over \sqrt{2} + 1}\bigg)$$

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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.

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