Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$
Edit:
This is what I got so far: $\int_{0}^{a}\sqrt{1+\left(\left(\sqrt{x}\right)'\right)^{2}}dx=\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\left[\begin{array}{cc}
t^{2}=1+\frac{1}{4x} & \sqrt{1+\frac{1}{4N}},\sqrt{1+\frac{1}{4a}}\\
2tdt=-\frac{1}{8x^{2}}dx & x=\frac{1}{4\left(t^{2}-1\right)}
\end{array}\right]=\\\lim_{N\to0}-\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\frac{t^{2}dt}{\left(t^{2}-1\right)^{2}}=\lim_{N\to0}\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\left(\frac{1}{4\left(t+1\right)}-\frac{1}{4\left(t+1\right)^{2}}-\frac{1}{4\left(t-1\right)}-\frac{1}{4\left(t-1\right)^{2}}\right)dt=\\=\lim_{N\to0}\frac{1}{4}\left[\ln\left(t+1\right)+\frac{1}{t+1}-\ln\left(t-1\right)+\frac{1}{t-1}\right]_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4a}}}{1+\frac{1}{4a}-1}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4N}}}{1+\frac{1}{4N}-1}\right]=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-8N\sqrt{1+\frac{1}{4N}}\right]=\\=\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}-0+0=\ln\left(4a\left(\sqrt{\frac{1}{a}+4}+2\right)+1\right)-8a\sqrt{1+\frac{1}{4a}}$
But it doesn't seem right... any Ideas what went wrong?
 A: If $u^2=1+\frac1{4x}$ then $x=\frac1{4(u^2-1)}$ and $\mathrm{d}x=-\frac{u}{2(u^2-1)^2}\,\mathrm{d}u$
$$
\begin{align}
\int_0^a\sqrt{1+\frac1{4x}}\,\mathrm{d}x
&=\frac12\int_{\sqrt{1+\frac1{4a}}}^\infty\frac{u^2}{(u^2-1)^2}\,\mathrm{d}u\\
&=\frac18\int_{\sqrt{1+\frac1{4a}}}^\infty\left(\frac1{(u-1)^2}+\frac1{u-1}+\frac1{(u+1)^2}-\frac1{u+1}\right)\mathrm{d}u\\
&=\frac18\left[-\frac1{u-1}+\log(u-1)-\frac1{u+1}-\log(u+1)\right]_{\sqrt{1+\frac1{4a}}}^\infty\\
&=\frac18\left[\frac{2u}{u^2-1}+\log\left(\frac{(u+1)^2}{u^2-1}\right)\right]_{u=\sqrt{1+\frac1{4a}}}\\
&=\frac14\left(\sqrt{4a(4a+1)}+\log\left(\sqrt{4a}+\sqrt{4a+1}\right)\right)
\end{align}
$$
A: You need to make a first substitution:
let u=  1+1/4x
yhen you will have to integrate: sqrt(1+u^2)du which is easy.
at the end you will get a formula with ln, tan inverse, and sec.
Hope it works
A: We have $$\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\frac{1}{4}\int_{0}^{a}\left(\frac{1}{\sqrt{x}\sqrt{1+4x}}+\frac{1+8x}{\sqrt{x}\sqrt{1+4x}}\right)dx
 $$ for the first integral use the substitution $4x=t^{2}
 $ (the second is trivial) to get $$\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\frac{1}{4}\left(\textrm{arcsinh}\left(2\sqrt{a}\right)+2\sqrt{a}\sqrt{1+4a}\right).
 $$
A: Hint: 
Use differential binomial. Note that $$\sqrt{1+\frac{1}{4x}}=0.5x^{-0.5}(1+4x)^{0.5}$$
