Question:
If the arc length formula of a function $f$ on an interval $[a,b]$ is given by $L_a^b=\int_a^b \sqrt{1+[f'(x)]^2} \ dx$. Find the arc length of $f(x)=4x^\frac{1}{2}+9$ on $[0,1]$.
We have $[f'(x)]^2=[2x^\frac{-1}{2}]^2=\frac{4}{x}$. Then,
$$L_0^1=\int_0^1 \sqrt{1+\frac{4}{x}} \ dx=\int_0^1 \sqrt{\frac{x+4}{x}} \ dx = \int_0^1 \frac{\sqrt{x+4}}{\sqrt{x}} \ dx=\int_0^1 (x+4)^\frac{1}{2} \ x^\frac{-1}{2} dx$$
I am stuck at this integral, I tried change of variable but it didn't work. Any suggestions? Thanks!
Best Answer
Let $u = \sqrt{1 + \frac{4}{x}}$, $x= \frac{4}{u^2-1}$
$\frac{du}{dx} = \frac{-2}{x^2 \cdot\sqrt{1 + \frac{4}{x}} }$
$dx = - \frac{x^2 \cdot\sqrt{1 + \frac{4}{x}}} {2}du =- \frac{8u}{(u^2-1)^2} du$
$\sqrt{1 + \frac{4}{x}} dx = - \frac{8u^2}{(u^2-1)^2} du$
$I=\int\limits_{\sqrt{5}}^{\infty} \frac{8u^2}{(u^2-1)^2} du$
Then just use partial fraction to compute this integral.