Find the length of the curve $x^2$ from $[1, 3]$

Question

I'm not getting the correct answer?

The problem asks to find the length of the curve $f(x)=x^2$ on the interval $[1,3]$. Using the definition of Arc Length = $$s=\int_1^3\left(\sqrt{1+[f'(x)]^2}\right)dx$$
Substitute $f'(x)= 2x$ $$s=\int_1^3\left(\sqrt{1+[2x]^2}\right)dx$$ Then u-substitute for $2x$.
$u=2x$ $\frac{du}{2}=dx$

$$s=\frac12\int_2^6\left(\sqrt{1+[u]^2}\right)du$$

From here I decided to use the Pythagorean Idenitie: $sec^2(\theta)-tan^2(\theta)=1$
$sec(\theta)=\sqrt{1^2+tan^2(\theta)}$
if $tan(\theta)= u$,
then,
$sec(\theta)=\sqrt{1+u^2}$

$$s=\frac12\int_2^6\left(sec(\theta)\right)du$$ convert $du$ to $d\theta$

$sec(\theta)=\sqrt{1+u^2}$
$sec(\theta)tan(\theta)d\theta=\frac{u}{\sqrt{1+u^2}}du$
$sec(\theta)tan(\theta)d\theta=\frac{tan(\theta)}{sec(\theta)}du$
$du=sec^2(\theta)d\theta$

$$s=\frac12\int_2^6(sec^3(\theta))d\theta$$

Then rewrite it for integration by parts

$$s=\frac12\int_2^6(sec(\theta)sec^2(\theta))d\theta$$

$\int(\frac{sec(\theta)}{2})sec^2(\theta)d\theta=\int(u)dv$
$u=\frac{sec(\theta)}{2}$
$du=\frac{sec(\theta)tan(\theta)}{2}d(\theta)$

$dv=sec^2(\theta)d\theta$
$v=tan(\theta)$

Integration by parts formula.

$$\int(u)dv=uv-\int(v)du$$

$$\int(\frac{sec^3(\theta)}{2})d\theta=\frac{sec(\theta)tan(\theta)}{2}-\int(\frac{tan^2(\theta)sec(\theta)}{2})d\theta$$

Substitute (sec$^2$($\theta$)-1) for tan$^2$($\theta$)

$$\int(\frac{sec^3(\theta)}{2})d\theta=\frac{sec(\theta)tan(\theta)}{2}-\int(\frac{(sec^2(\theta)-1)sec(\theta)}{2})d\theta$$

Distribute sec($\theta$)

$$\int(\frac{sec^3(\theta)}{2})d\theta=\frac{sec(\theta)tan(\theta)}{2}-\int(\frac{(sec^3(\theta)-sec(\theta)}{2})d\theta$$

Seperate into two integrals

$$\int(\frac{sec^3(\theta)}{2})d\theta=\frac{sec(\theta)tan(\theta)}{2}-\int(\frac{sec^3(\theta)}{2})d\theta+\int(\frac{sec(\theta)}{2})d\theta$$

Collect like integrals.

$$2\int(\frac{sec^3(\theta)}{2})d\theta=\frac{sec(\theta)tan(\theta)}{2}+\int(\frac{sec(\theta)}{2})d\theta$$

Pull out coefficient and Integrate sec($\theta$)

$$2\int(\frac{sec^3(\theta)}{2})d\theta=\frac{sec(\theta)tan(\theta)}{2}+\frac12 ln\lvert sec(\theta)+tan(\theta)\rvert$$

Divide by 2 on both sides of the equality sign

$$\int_2^6(\frac{sec^3(\theta)}{2})d\theta=\left[\frac{sec(\theta)tan(\theta)}{4}+\frac14 (ln\lvert sec(\theta)+tan(\theta)\rvert)\right]_2^6$$

Substitute from trig to algebraic

$$ \int_2^6(\frac {sec^3(\theta)}{2}) d\theta = \left[ \frac{(\sqrt{1+u^2})(u)}{4}+\frac14 (ln\lvert (\sqrt{1+u^2})+(u)\rvert) \right]_2^6 $$

$$ \left[ \frac{(\sqrt{1+(6)^2})(6)}{4}+\frac14 (ln\lvert (\sqrt{1+(6)^2})+(6)\rvert) \right]-\left[ \frac{(\sqrt{1+(2)^2})(2)}{4}+\frac14 (ln\lvert (\sqrt{1+(2)^2})+(2)\rvert) \right] $$

$$ \bbox[5px,border:2px solid red] {\left[ \frac{(3\sqrt{37})}{2}+\frac14 (ln\lvert (\sqrt{37})+6\rvert) \right]-\left[ \frac{\sqrt{5}}{2}+\frac14 (ln\lvert (\sqrt{5})+2\rvert) \right]} $$

$$ \left[ (9.12)+(0.62) \right]-\left[ (1.12)+(0.36) \right] $$

$$\bbox[5px,border:2px solid red]{\approx 8.0061098067}$$

Use the Distance Formula to approximate and double check answer. $$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ $(1,1)and(3,9)$

$$\sqrt{68}$$

$$\approx 8.24621125124$$

Answer 1

The exact expression you have obtained in the penultimate red box is correct, but the final numerical value in the last red box is not correct, and should be $8.268145901064$

When you do the substitution $2x=\sinh u$ this leads to $$L=\frac 12\int \cosh^2 udu=\frac 14(u+\sinh u\cosh u)$$

So when you apply the limits you are evaluating $$\frac 14(\operatorname{arsinh}6+6\sqrt{37}-\operatorname{arsinh}2-2\sqrt{5})$$