Answer
$\frac{2\sqrt 5+ln(2+\sqrt5)}{4}$
Work Step by Step
Step 1: The arc length L=$\int^{1}_{0}$$\sqrt{1+(\frac{dy}{dx})^{2}}$dx
In this problem, y=$x^{2}$; so, $\frac{dy}{dx}$=2x
$\int^{1}_{0}$$\sqrt{1+(\frac{dy}{dx})^{2}}$dx=$\int^{1}_{0}$$\sqrt{1+(2x)^{2}}$dx
Step 2: Now to evaluate this integral, we need to use trigonometry substitution method with the trigonometry relationship shown by the image below. $\sqrt{1+(2x)^{2}}$ is replaced by sec$\theta$ and dx is replaced by $\frac{1}{2}$$sec^{2}$$\theta$d$\theta$
$\int^{1}_{0}$$\sqrt{1+(2x)^{2}}$dx= $\frac{1}{2}$$\int^{}_{}$sec$\theta$($sec^{2}$$\theta$)d$\theta$=$\frac{1}{2}$$\int^{}_{}$$sec^{3}$$\theta$d$\theta$
Step 3: Here to evaluate $\int$$sec^{3}$$\theta$d$\theta$, let u=sec$\theta$ & dv= $sec^{2}$$\theta$d$\theta$. So, v=$\int$$sec^{2}$$\theta$d$\theta$=tan$\theta$
$\int$$sec^{3}$$\theta$d$\theta$=sec$\theta$·tan$\theta$-$\int$tan$\theta$(sec$\theta$·tan$\theta$)d$\theta$=sec$\theta$·tan$\theta$-$\int$$tan^{2}$$\theta$sec$\theta$d$\theta$
Then, we use the identity $tan^{2}$$\theta$= $sec^{2}$$\theta$-1
So $\int$$sec^{3}$$\theta$d$\theta$=sec$\theta$·tan$\theta$-$\int$($sec^{2}$$\theta$-1)sec$\theta$d$\theta$
When we split this integral, we would have $\int$$sec^{3}$$\theta$d$\theta$=sec$\theta$·tan$\theta$-$\int$$sec^{3}$$\theta$d$\theta$+$\int$sec$\theta$d$\theta$
We find we have just produced an equation about $\int$$sec^{3}$$\theta$d$\theta$, by solving it we have:
$\int$$sec^{3}$$\theta$d$\theta$=$\frac{1}{2}$(sec$\theta$·tan$\theta$+ $\int$sec$\theta$d$\theta$)=$\frac{1}{2}$(sec$\theta$·tan$\theta$+ln|tan$\theta$+sec$\theta$|)
Step 4: Let's plug in $\int$$sec^{3}$$\theta$d$\theta$=$\frac{1}{2}$(sec$\theta$·tan$\theta$+ln|tan$\theta$+sec$\theta$|) into
$\int^{1}_{0}$$\sqrt{1+(2x)^{2}}$dx=$\frac{1}{2}$$\int^{}_{}$$sec^{3}$$\theta$d$\theta$ we got from step 2, and we have:
$\int^{1}_{0}$$\sqrt{1+(2x)^{2}}$dx=$\frac{1}{4}$(sec$\theta$·tan$\theta$+ln|tan$\theta$+sec$\theta$|)
Then, we convert $\theta$ back to x using the trigonometry relationship shown by the image below: $\frac{1}{4}$(sec$\theta$·tan$\theta$+ln|tan$\theta$+sec$\theta$|) = $\frac{1}{4}$[$\sqrt{1+(2x)^{2}}$ (2x)+ln|2x+$\sqrt{1+(2x)^{2}}$|]
Finally, let's plug $x_{f}$=1 & $x_{i}$=0 into this expression for $\int^{1}_{0}$$\sqrt{1+(2x)^{2}}$dx, and we get the final answer, $\frac{2\sqrt 5+ln(2+\sqrt5)}{4}$.
