Calculating arc-length using integrals

Question

Find the length of the following curves:

A) $y=(2x^2 -ln(x))/4$ with $2 \lt x \lt 4$

B) $y=ln(x)$ with $1 \lt x \lt \sqrt{3}$

Well, I've used the formula $\int_a^b \sqrt{1+(f'(x))^2} \,dx$ to calculate the arc lenght. Now, I have some problems when integrating.

For A), I get $\int_2^4 \sqrt{1+(\frac{4x^2 -1}{2x})^2} \,dx$. I tried calling $\frac{4x^2 -1}{2x}$=$u$ so that I can use $cosh(x)^2=1+sinh(x)$ but I can't do that because of $dx$.

As for B), I get $\int_1^\sqrt{3} \sqrt{1+(\frac{1}{x})^2} \,dx$, and I have the same idea and problem, I tried using $cosh(x)^2=1+sinh(x)$ but $dx$ is my problem.

Also, why does the activity say $2 \lt x \lt 4$ and not $2 \le x \le 4$? Does it change anything?

Answer 1

Hint: For the second $$\int \sqrt{1+(\frac{1}{x})^2} \,dx=\int \frac{\sqrt{1+x^2}}{x^2}\ x dx$$ let substitution $x^2+1=u^2$ then you have $xdx=udu$ by differentiation thus the integral is $$\int \dfrac{u^2}{u^2-1}du=\int 1+\dfrac12\dfrac{1}{u-1} - \dfrac12\dfrac{1}{u+1}\ du$$

Answer 2

Hint

For the first one, after expanding, you have $$\sqrt{x^2+\frac{1}{16 x^2}+\frac{1}{2}}=\sqrt{\left(x+\frac 1 {4x}\right)^2}=|x|+\frac 1 {4|x|}$$

$$\int\sqrt{x^2+\frac{1}{16 x^2}+\frac{1}{2}}\,dx=\dfrac{\ln\left(\left|x\right|\right)+2x^2}{4}$$