Integral $\sqrt{1+\frac{1}{4x}}$

Question

$$\mathbf\int\sqrt{1+\frac{1}{4x}} \, dx$$

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?

Answer 1

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.

Answer 2

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 $$dx=-\frac{t}{(t^2-1)^2}dt$$

Answer 3

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: $u=\cfrac{1}{4x}, r=\sqrt{u+1}$, $$=\cfrac{1}{4}\bigg[4x\sqrt{1+\cfrac{1}{4x}} + arctanh\bigg(\sqrt{\cfrac{1}{4x}+1}\bigg)\bigg]+C$$

Answer 4

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*}

Answer 5

$$I=\int\frac{\sqrt{4x+1}}{2\sqrt{x}}dx$$ 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 $$\int\sqrt{x^2+a^2}=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}ln|x+\sqrt{x^2+a^2}|$$