How to get rid of the integral in this equation $\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dx}f(x)\right)^2}dx}$? How to get rid of the integral $\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dx}f(x)\right)^2}dx}$ when $f(x)=x^2$?
 A: $$\int_{x_0}^x\!\!\!\sqrt{1+\left(\frac{d}{d\xi}f(\xi)\right)^2}\,d\xi\quad\text{with}\quad f(x)=x^2$$
$$ = \int_{x_0}^x\!\!\!\sqrt{1+4\xi^2}\,d\xi $$
Substituting $2\xi=\sinh(u) \Rightarrow d\xi = \frac{1}{2}\cosh(u)\,du$ results in
$$ \int\!\frac{1}{2}\cosh(u)\sqrt{1+\sinh^2(u)}\,du $$
$$ = \frac{1}{2}\int \cosh^2(u)\, du $$

Solving this integral with partial integration gives
  $$ \int_a^b \cosh^2(u)\,du = \left[ \cosh(u)\sinh(u) \right]_{u=a}^{b} - \int_a^b \sinh^2(u)\, du $$
  $$ = \left[ \cosh(u)\sinh(u) \right]_{u=a}^{b} - \int_a^b \left(\cosh^2(u)-1 \right)\,du $$
  $$ \Rightarrow \int_a^b \cosh^2(u)\,du = \frac{1}{2}\left(\left[ \cosh(u)\sinh(u) \right]_{u=a}^{b} + (b-a)\right) $$

Pluging in this solution and subsequently undoing the substitution (so we can keep the old limits) gives
$$ \frac{1}{4} \left(\left[ 2\xi\sqrt{1+4\xi^2} \right]_{\xi=x_0}^{x} + (\sinh^{-1}(2x)-\sinh^{-1}(2x_0))\right) $$
A: As in other answers, let $2x=\tan(\theta)$
$$
\begin{align}
\int\sqrt{1+4x^2}\mathrm{d}x
&=\frac12\int\sec(\theta)\,\mathrm{d}\tan(\theta)\\
&=\frac12\int\sec^4(\theta)\,\mathrm{d}\sin(\theta)\\
&=\frac12\int\frac{\mathrm{d}\sin(\theta)}{(1-\sin^2(\theta))^2}\\
&=\frac18\int\left({\small\frac{1}{1-\sin(\theta)}+\frac{1}{(1-\sin(\theta))^2}+\frac{1}{1+\sin(\theta)}+\frac{1}{(1+\sin(\theta))^2}}\right)\mathrm{d}\sin(\theta)\\
&=\frac18\left(\log\left(\frac{1+\sin(\theta)}{1-\sin(\theta)}\right)+\frac{1}{1-\sin(\theta)}-\frac{1}{1+\sin(\theta)}\right)+C\\
&=\frac14\left(\log\left(\tan(\theta)+\sec(\theta)\right)+\tan(\theta)\sec(\theta)\right)+C\\
&=\frac14\log\left(2x+\sqrt{1+4x^2}\right)+\frac12x\sqrt{1+4x^2}+C\tag{1}
\end{align}
$$
Then take the difference of $(1)$ at the limits of integration.
A: Summarising the comments, you'll get
$$
\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dt}f(t)\right)^2}dt}
=\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dt}t^2\right)^2}dt}
=\int\limits_{x_0}^{x}{\sqrt{1+4t^2}dt}
$$
To solve the last one substitute $t=\tan(u)/2$ and $dt=\sec^2(u)/2du$. Then $\sqrt{1+4t^2}= \sqrt{\tan^2(u)+1}=\sec(u)$, so we get as antiderviative:
$$
\begin{eqnarray}
\frac{1}{2}\int \sec^3(u) du&=&\frac{1}{4}\tan(u)\sec(u)+\frac{1}{4}\int \sec(u)du+\text{const.}\\
&=&\frac{1}{4}\tan(u)\sec(u)+\frac{1}{4}\log(\tan(u)+\sec(u))+\text{const.} \\
&=& \frac{t}{2}\sqrt{1+4t^2}+\frac{1}{4}\log(2t+\sqrt{1+4t^2})+\text{const.}\\
&=& \frac{1}{4}\left(2t\sqrt{1+4t^2}+\sinh^{-1}(2t) \right)+\text{const.}.
\end{eqnarray}
$$
Put in your limits and your done:
$$
\int\limits_{x_0}^{x}{\sqrt{1+4t^2}dx}=\left[\frac{1}{4}\left(2t\sqrt{1+4t^2}+\sinh^{-1}(2t) \right) \right]_{x_0}^x
$$
A: Go back to the basics!
You need a derivative of the curve to begin with.
If the integrand has root of sum of squares in the denom
Immediately draw a right triangle with appropriate sides and the rest is an algebraic walk.
