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?

• 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

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 $x=2$.

• 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?

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 $[0,2]$. • You mean $x=\cfrac{y^2}{4}$. – Cameron Buie May 28 '12 at 14:44
• @Cameron: Thanks for that - – davidlowryduda 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 $$\int_\infty^2-\sqrt{u}\,\left(-\frac{du}{(u-1)^2}\right)=\int_2^\infty\frac{\sqrt{u}}{(u-1)^2}\,du$$

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*}

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.

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

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}

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.

• (The link you had was not working, so I updated it.) – Andrés E. Caicedo Aug 4 '18 at 14:29