Answer
ln($\sqrt 10$+$\sqrt 5$-$\sqrt 2$-1)-ln2+$\sqrt 5$-$\sqrt 2$
Work Step by Step
The function y=lnx
Arc length L=$\int^{2}_{1}${√[1+$(\frac{dy}{dx})^{2}$]} dx
=$\int^{2}_{1}${√[1+$(\frac{1}{x})^{2}$]} dx
=$\int^{2}_{1}$$\frac{√(x^{2}+1)}{√(x^{2})}$dx
=$\int^{2}_{1}$$\frac{√(x^{2}+1)}{x}$dx (As x∈[1,2] is positive)
Now, ($x^{2}$+1) in the numerator of our integrand has the form of ($x^{2}$+$a^{2}$), so we can try to substitute x with a·tanθ so that $x^{2}$+$a^{2}$=$a^{2}$$tan^{2}θ$+$a^{2}$=$a^{2}$(1+$tan^{2}θ$)=$a^{2}$$sec^{2}θ$
In this problem, a=1. So let x=tanθ, and dx=$sec^{2}θ$dθ.
$\int^{2}_{1}$$\frac{√(x^{2}+1)}{x}$dx =$\int^{}_{}$$\frac{√(tan^{2}θ+1)}{tanθ}$($sec^{2}θ$dθ)
= $\int^{}_{}$$\frac{√(sec^{2}θ)}{tanθ}$($sec^{2}θ$dθ)
Now let's assume θ to be an acute angle, $secθ$>0, so √($sec^{2}θ$)=$secθ$
The expression = $\int^{}_{}$$\frac{(secθ)sec^{2}θ}{tanθ}$dθ
Use the identity $sec^{2}θ$=1+$tan^{2}θ$, the expression=$\int^{}_{}$$\frac{(secθ)(1+tan^{2}θ)}{tanθ}$dθ
Then we spit the integral and simplify it, the expression=$\int^{}_{}$$\frac{secθ}{tanθ}$dθ+$\int^{}_{}$secθ·tanθdθ
$\int^{}_{}$$\frac{secθ}{tanθ}$dθ=$\int^{}_{}$$\frac{1}{tanθ·cosθ}$dθ=$\int^{}_{}$$\frac{1}{sinθ}$dθ=$\int^{}_{}$$cscθ$dθ=ln|cotθ-cscθ|
& $\int^{}_{}$secθ·tanθdθ=secθ
So the result of our integral = ln|cotθ-cscθ|+secθ
Finally, we convert θ back to x based on the trigonometry relationship shown by the image below:
ln|cotθ-cscθ|+secθ=ln |$\frac{1}{x}$-$\frac{√(x^{2}+1)}{x}$|+√($x^{2}$+1)
When we plug in $x_{f}$=2, $x_{i}$=1, we have the final answer: ln($\sqrt 10$+$\sqrt 5$-$\sqrt 2$-1)-ln2+$\sqrt 5$-$\sqrt 2$
