Find the length of $\sqrt x$. Let $f(x) = \sqrt x$. Find the length of the curve for $0\le x \le a$.
So I know the formula is:
$$\int_0^a \sqrt{(1+(f')^2(x)} \ dx = \ldots = \int_0^a \sqrt{1+\frac{1}{4x}} \ dx$$
Now, how do I evaulate this integral? 
I tried the subtitution $t = \sqrt {1+\frac{1}{4x}}$ but it became somewhat complicated.
 A: $$\int\sqrt{1+\frac{1}{4x}}\space\text{d}x=$$
$$\int\sqrt{1+\frac{x+\frac{1}{4}}{x}}\space\text{d}x=$$

Substitute $u=\frac{x+\frac{1}{4}}{x}$ and $\text{d}u=\left(\frac{1}{x}-\frac{x+\frac{1}{4}}{x^2}\right)\space\text{d}x$:

$$-\frac{1}{4}\int\frac{\sqrt{u}}{(1+u)^2}\space\text{d}u=$$

Substitute $s=\sqrt{u}$ and $\text{d}s=\frac{1}{2\sqrt{u}}\space\text{d}u$:

$$-\frac{1}{2}\int\frac{s^2}{(1-s^2)^2}\space\text{d}s=$$
$$-\frac{1}{2}\int\frac{s^2}{(s^2-1)^2}\space\text{d}s=$$
$$-\frac{1}{2}\int\left(-\frac{1}{4(s+1)}+\frac{1}{4(s+1)^2}+\frac{1}{4(s-1)}+\frac{1}{4(s-1)^2}\right)\space\text{d}s=$$
$$\frac{1}{8}\int\frac{1}{s+1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s+1)^2}\space\text{d}s-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$

Substitute $p=s+1$ and $\text{d}p=\text{d}s$:

$$\frac{1}{8}\int\frac{1}{p}\space\text{d}p-\frac{1}{8}\int\frac{1}{(s+1)^2}\space\text{d}s-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
$$\frac{\ln|p|}{8}-\frac{1}{8}\int\frac{1}{(s+1)^2}\space\text{d}s-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$

Substitute $w=s+1$ and $\text{d}w=\text{d}s$:

$$\frac{\ln|p|}{8}-\frac{1}{8}\int\frac{1}{w^2}\space\text{d}w-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$

Substitute $v=s-1$ and $\text{d}v=\text{d}s$:

$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{1}{8}\int\frac{1}{v}\space\text{d}v-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{\ln|v|}{8}-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$

Substitute $q=s-1$ and $\text{d}q=\text{d}s$:

$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{\ln|v|}{8}-\frac{1}{8}\int\frac{1}{q^2}\space\text{d}q=$$
$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{\ln|v|}{8}+\frac{1}{8q}+\text{C}=$$
$$\frac{\ln|s+1|}{8}+\frac{1}{8(s+1)}-\frac{\ln|s-1|}{8}+\frac{1}{8(s-1)}+\text{C}=$$
$$\frac{\ln\left|\sqrt{u}+1\right|}{8}+\frac{1}{8\left(\sqrt{u}+1\right)}-\frac{\ln\left|\sqrt{u}-1\right|}{8}+\frac{1}{8\left(\sqrt{u}-1\right)}+\text{C}=$$
$$\frac{\ln\left|\sqrt{\frac{x+\frac{1}{4}}{x}}+1\right|}{8}+\frac{1}{8\left(\sqrt{\frac{x+\frac{1}{4}}{x}}+1\right)}-\frac{\ln\left|\sqrt{\frac{x+\frac{1}{4}}{x}}-1\right|}{8}+\frac{1}{8\left(\sqrt{\frac{x+\frac{1}{4}}{x}}-1\right)}+\text{C}$$
A: You could substitute $x=y^2,\ dx=2y\,dy$ to transform your integral into
$$L=\int_0^{\sqrt a} \sqrt{4y^2+1}\,dy$$
then use the first equation from Wikipedia's List of integrals of irrational functions.
A: Substitute $t = \ln(8x + 1 + \sqrt{4x^2 + x})$, then you have ${\rm d}x = \frac18(e^t-e^{-t})(e^t+e^{-t}){\rm d}t$, $x = \frac{(e^t - e^{-t})^2}{16}$ and $\sqrt{4x^2+x} = \frac{(e^t-e^{-t})(e^t+e^{-t})}{8}$
