Integral $\sqrt{1+\frac{1}{4x}}$ $$\mathbf\int\sqrt{1+\frac{1}{4x}} \, dx$$
This integral came up while doing an arc length problem and out of curiosity I typed it into my TI 89 and got this output
$$\frac{-\ln(|x|)+2\ln(\sqrt\frac{4x+1}{x}-2)-2x\sqrt\frac{4x+1}{x}}{8}\ $$
How would one get this answer by integrating manually?
 A: Sub $x=u^2$. Then the integral is equal to
$$\int du \, \sqrt{1+4 u^2} $$
Now set $u = \frac12 \sinh{v}$; now the integral is equal to
$$\frac12 \int dv \, \cosh^2{v} = \frac14 \int dv \, (1+\cosh{2 v})= \frac{v}{4} + \frac14 \sinh{v} \cosh{v} + C$$
Sub back $v=\sinh^{-1}{(2 \sqrt{x})} = \log{\left (2 \sqrt{x} + \sqrt{1+4 x} \right)}$ and you are done.
A: Set $u=\cfrac{1}{4x}, du=\cfrac{-1}{4x^2}$. Substitute:
$$\int \sqrt{1+u}(-4x^2)du$$
Since $4x=\cfrac{1}{u}, x=\cfrac{1}{4u}$,
$$=\int \cfrac{-1}{4u^2}\sqrt{1+u}du$$
Integrate by parts: $s=\sqrt{1+u}, ds=\cfrac{1}{2\sqrt{1+u}}du, dt=\cfrac{1}{u^2}du, t=\cfrac{-1}{u}$,
$$=\cfrac{-1}{4}\bigg[\cfrac{-\sqrt{1+u}}{u} - \int\cfrac{-1}{u}\cfrac{1}{2\sqrt{1+u}}du\bigg]$$
Substitute again, $r=\sqrt{u+1}, dr=\cfrac{1}{2\sqrt{u+1}}du=\cfrac{1}{2r}du \rightarrow 2rdr=du$
$$=\cfrac{-1}{4}\bigg[\cfrac{-\sqrt{1+u}}{u} - \int\cfrac{-1}{u}\cfrac{1}{2r}2rdr\bigg]$$
$$=\cfrac{-1}{4}\bigg[\cfrac{-\sqrt{1+u}}{u} - \int\cfrac{1}{1-r^2}dr\bigg]$$
Use the identity $\int \cfrac{1}{1-x^2} dx = arctanh(x)+C$.
$$=\cfrac{-1}{4}\bigg[\cfrac{-\sqrt{1+u}}{u} - arctanh(r) + C\bigg]$$
Substitute back in: $u=\cfrac{1}{4x}, r=\sqrt{u+1}$,
$$=\cfrac{1}{4}\bigg[4x\sqrt{1+\cfrac{1}{4x}} + arctanh\bigg(\sqrt{\cfrac{1}{4x}+1}\bigg)\bigg]+C$$
A: Assume $x>0.$ Let $u=\sqrt{x},$ then $u^{2}=x,$ $dx=2udu,$ and $1+\frac{1}{4x%
}=1+\frac{1}{4u^{2}}$
\begin{equation*}
\int \sqrt{1+\frac{1}{4x}}dx=\int \sqrt{1+\frac{1}{4u^{2}}}(2u)du=\int \sqrt{%
(2u)^{2}+1}du.
\end{equation*}
Now, substitution trigonometric, $2u=\tan \theta ,$ $du=\frac{1}{2}\sec
^{2}\theta d\theta ,$ and $\sqrt{(2u)^{2}+1}=\sec \theta ,$ hence
\begin{equation*}
\int \sqrt{(2u)^{2}+1}du=\int (\sec \theta )\frac{1}{2}\sec ^{2}\theta
d\theta =\frac{1}{2}\int \sec ^{3}\theta d\theta .
\end{equation*}
Then, we use the well-known integral
\begin{equation*}
\int \sec ^{3}\theta d\theta =\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}%
\ln \left\vert \sec \theta +\tan \theta \right\vert +2C
\end{equation*}
it follows that
\begin{eqnarray*}
\int \sqrt{1+\frac{1}{4x}}dx &=&\frac{1}{4}\sec \theta \tan \theta +\frac{1}{%
4}\ln \left\vert \sec \theta +\tan \theta \right\vert +C \\
&=&\frac{1}{4}(\sqrt{(2u)^{2}+1})(2u)+\frac{1}{4}\ln \left\vert \sqrt{%
(2u)^{2}+1}+2u\right\vert +C \\
&=&\frac{1}{2}u\sqrt{1+4u^{2}}+\frac{1}{4}\ln \left\vert \sqrt{(2u)^{2}+1}%
+2u\right\vert +C \\
&=&\frac{1}{2}\sqrt{x}\sqrt{1+4x}+\frac{1}{4}\ln \left\vert \sqrt{1+4x}+2%
\sqrt{x}\right\vert +C.\ \ \ \ \blacksquare 
\end{eqnarray*}
A: $$I=\int\frac{\sqrt{4x+1}}{2\sqrt{x}}dx$$   Use $$\sqrt{x}=t$$ $\implies$ $$\frac{dx}{2\sqrt{x}}=dt$$
we get  $$I=\int \sqrt{4t^2+1}dt=2\int\sqrt{t^2+0.5^2}dt$$
now use standard integral $$\int\sqrt{x^2+a^2}=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}ln|x+\sqrt{x^2+a^2}|$$
A: setting $$t=\sqrt{1+\frac{1}{4x}}$$ we get $$\int-\frac{t^2}{(t^2-1)^2}dt$$
you will get $$x=\frac{1}{4(t^2-1)}$$ and $$dx=-\frac{t}{(t^2-1)^2}dt$$
