Calculating the length of a curve I must calculate the length of the curve $$\frac{x^2}{72} - 9 \ln(x)$$ for $$9 \leq x \leq 9 e.$$
We know that formula för length is to integrate $$\sqrt{1+f'^2(x)}dx$$ between the two points of interest.
I calculated the derivata to be:
$$\frac{x}{36} - \frac{9}{x}$$
And squared:
$$\frac{x^2}{36^2} - 2 \cdot \frac{x}{36}\cdot\frac{9}{x} + \frac{9^2}{x^2} = $$
$$\frac{x^2}{36^2} - \frac{1}2 + \frac{9^2}{x^2}$$
The integral is then:
$$\int_9^{9e} \sqrt{1+\frac{x^2}{36^2} - \frac{1}2+ \frac{9^2}{x^2}} dx$$
$$\int_9^{9e} \sqrt{\dfrac{36^2x^2+x^4- 748x^2+ 9^2 \cdot 36^2}{36^2 \cdot x^2}} dx$$
$$\int_9^{9e} \sqrt{\dfrac{x^4 + 748x^2+ 9^2 \cdot 36^2}{36^2 \cdot x^2}} dx.$$
And I have no idea how to integrate that. The examples we have in the course is with trigonometric functions with a replacement with $$\tan^2(x)$$ but it is not applicable here...
 A: This is a classic trick:  They have the middle term equal to $-1/2$ so that when you add $1$, it becomes $+1/2$.  So the original thing you square is of the form $(a-b)^2$ and the factorization of the new expression is $(a+b)^2$.  
In other words, the expression inside your radical factors as
$$\left(\frac{x}{36} + \frac{9}{x}\right)^2.$$
Take the square root of that and you're golden.
Edit:  Book arclength problems are often contrived so that under the square root you have the form $(a-b)^2+4ab$.  If you simplify this you get $a^2-2ab+b^2+4ab =a^2+2ab+b^2 = (a+b)^2$.  The difference between $(a-b)^2$ and $(a+b)^2$ is only the sign on the $2ab$ term.  By adding $4ab$, you just change the sign on $2ab$, so you're just changing the expression from $(a-b)^2$ to $(a+b)^2$.  
In your problem, $a=x/36$ and $b+9/x$, so that $4ab = 1$.  Your calculation should look like
$$\int \sqrt{1 + \left(\frac{x}{36}-\frac{9}{x}\right)^2} \; dx 
 = \int\sqrt{ 1+\frac{x^2}{36^2}-\frac{1}{2} + \frac{81}{x^2}} \; dx.$$
Now adding the $1$ to the $-1/2$ really just changes the sign on the $1/2$, so we know how to factor the remaining expression:
$$=\int\sqrt{ \frac{x^2}{36^2}+\frac{1}{2} + \frac{81}{x^2}} \; dx
 = \int \sqrt{\left(\frac{x}{36}+\frac{9}{x}\right)^2} \; dx 
 = \int \frac{x}{36}-\frac{9}{x} \; dx. $$
A: There is a nice substitution.
Let $$\frac{x}{36} - \frac{9}{x}=t \implies x=18 \left(\sqrt{t^2+1}+t\right)\implies dx=18 \left(\frac{t}{\sqrt{t^2+1}}+1\right)$$ which make
$$\int \sqrt{1+\left(\frac{x}{36}-\frac{9}{x}\right)^2}\,dx=18\int \left(t+\sqrt{t^2+1}\right)\,dt$$ which seems to be easy to compute.
