<s>[INST] 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...
 [/INST] 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.
</s>