Evaluating ${\int_{0}^{1}\sqrt{1+\frac{1}{3x}}\text{ d}x}$ using $\int f^{-1}(x)dx=x f^{-1}(x)-F(f^{-1}(x))+C$ While studying, I countered problem $\text{#}2$ here. $\text{(UCHICAGO REU 2019 - MATH GRE PREP: WEEK 2)}$.
I saw this also. But wanted to know if my way is also correct (regardless of the time consuming).

$\color{red}{\text{Problem: Which of the following is closest to the value of this integral}}$
$$\color{red}{\int_{0}^{1}\sqrt{1+\frac{1}{3x}}\text{ d}x?}$$
$\color{red}{\text{(A) }1 \space\space\space\space\space\space\space\space\space\space\text{(B) }1.2 \space\space\space\space\space\space\space\space\space\space \color{blue}{\boxed{{\text{(D) }1.6}}}\space\space\space\space\space\space\space\space\space\space\text{(D) }2\space\space\space\space\space\space\space\space\space\space\text{(E) }\text{The integral doesn’t converge.} }$

My Attempt:
First, I found the inverse of $f^{-1}(x)=\sqrt{1+\frac{1}{3x}}$ as a separate problem. As usual, I replaced $x$ with $y$ and $y$ with $x$, then I solved for $y$. I got $f(x)=\frac{1}{3(x^2-1)}$, which (using partial fractions), $f(x)=\frac{1}{6}\bigg(\frac{1}{x-1}-\frac{1}{x+1}\bigg)$.
Second, I evaluated $F(x)=\int f(x)\text{ d}x=\frac{1}{6}\int \big(\frac{1}{x-1}-\frac{1}{x+1}\big)\text{ d}x=\frac{1}{6}\log\bigg(\frac{x-1}{x+1}\bigg)$ [constant of integration is not added in this step, because we have definite integral, actually].
Next, I evaluated $F(f^{-1}(x))=\frac{1}{6}\log\bigg(\frac{\sqrt{1+\frac{1}{3x}}-1}{\sqrt{1+\frac{1}{3x}}+1}\bigg)$.
Finally, I used the formula:
$$\int f^{-1}(x)dx=x f^{-1}(x)-F(f^{-1}(x))+C$$
So, $\int_{0}^{1}\sqrt{1+\frac{1}{3x}}\text{ d}x=x\sqrt{1+\frac{1}{3x}}-\frac{1}{6}\log\bigg(\frac{\sqrt{1+\frac{1}{3x}}-1}{\sqrt{1+\frac{1}{3x}}+1}\bigg)\bigg|_{x=0}^{x=1}$
Plugging the upper limit of integration, $x=1$, we get $\frac{2\sqrt{3}}{3}-\frac{1}{6}\log(7-4\sqrt{3})$

My problems:

*

*Even though this might not be an efficient way to save the time in GRE MATH SUBJECT TEST, I am still confused if my way was correct to solve such integral.


*If the way is correct, how can we evaluate it at the lower limit of integration, that is:
$$\lim_{x \rightarrow {0^+}}\bigg(x\sqrt{1+\frac{1}{3x}}-\frac{1}{6}\log\bigg(\frac{\sqrt{1+\frac{1}{3x}}-1}{\sqrt{1+\frac{1}{3x}}+1}\bigg)\bigg)$$
This should be $0$ as my calculator says. (By the way, calculators are prohibited in the test). I thought about applying L'Hospital's rule, but I did not success. I am not sure if this limit problem is harder than the original integral problem.

*

*How can we estimate $\frac{2\sqrt{3}}{3}-\frac{1}{6}\log(7-4\sqrt{3})$ to at least one decimal place? (No high accuracy is required as the given choices are far enough).


Your help would be appreciated. THANKS!
 A: Hm let's see. If we're only focused on approximating the integral, analytically solving it seems inefficient because in the end we do not seek such a thing, and as you can see, even if you have the closed form answer, it's useless if you cannot convert this into a decimal!
Instead let's use integral approximation methods. Our function decreases rapidly from infinity at $0$, before smoothing out to an asymptote. Hence, on the interval $(0,1]$, a good approximation method would be simpson's rule.
Since $0$ is a vertical asymptote, we'll use a left endpoint of $0.01$ for a close enough approximation. We'll use the 1/3 and 4-2-...-2-4 version of the rule.
Partition the interval into $4$ subintervals, to wit, $$[0.01, 0.25],\quad[0.25, 0.5],\quad[0.5, 0.75],\quad [0.75, 1]$$
First, establish $$f(x)=\sqrt{1+\frac1{3x}}$$
We have $f(0.01)\approx 5.86$, $4f(0.25)\approx 6.11$, $2f(0.5)\approx 2.58$, $4f(0.75)\approx 4.81$, $f(1)\approx 1.15$.
Lastly, $\frac{\Delta x}3\approx 0.0833$.
In the end, our integral is approximately $$0.0833(5.86+6.11+2.58+4.81+1.15)\approx 1.7$$
This is closest to $\boxed{\text{D}}$.
Arguably, this margin of error is still pretty large. It would be up to you during the test to determine the number of subintervals/accuracy you want to compute this to. imo $4$ subintervals is a pretty good approximation, but if you want to be more sure you could to $6$ or $8$. Depends on how much time you have to spare i guess.
A: I will concentrate only on the last two questions.
For the limit, you could simplify your expression. In the first term, move $x$ under the square root. For the expression in the log, multiply both numerator and denominator by $\sqrt{3x}$. Then you get $$\lim_{x\to 0^+}\left(\sqrt{x^2+\frac x3}-\frac16\log\frac{\sqrt{3x+1}-\sqrt{3x}}{\sqrt{3x+1}+\sqrt{3x}}\right)$$
Then you just plug in $x=0$ and you get $$\sqrt 0-\frac16\log\frac 11=0$$
For the last question, you can easily calculate $\sqrt3\approx 1.73$. Just keep two digits after decimal point. Then $$\frac{2\sqrt 3}3\approx 1.15$$For the logarithm, it would be easier if you rewrite it using $-\log x=\log\frac1x$. Therefore your expression is $$1.15+\frac16\log(13.92)$$
The log is between $2$ and $3$, so your result will be between $1.15+0.33=1.48$ and $1.15+0.5=1.65$.
