What is the integral of $\frac{\sqrt{x^2 +4}}{x}dx$

Question

I use trig substitution then get to this step but then I get stuck: $\int \frac{2\sec ^3\theta}{\tan \theta}d\theta$ anything I do seems to further complicate it. Thanks in advance.

Answer 1

HINT:

$$\frac{\sqrt{4+x^2}}x=\frac{x\sqrt{4+x^2}}{x^2}$$

Write $\sqrt{4+x^2}=u\implies4+x^2=u^2,x\ dx=u\ du$

$$\int\frac{\sqrt{4+x^2}}xdx=\int\frac u{u^2-4}u\ du=\int\frac{u^2-4+4}{u^2-4}du=\cdots$$

Answer 2

Hint: Rewrite the integrand just in terms of $\sin$ and $\cos$, and manipulate the expression so that one can substitute $u = \sin \theta$ or $u = \cos \theta$.

Write $$2 \int \frac{d\theta}{\sin \theta \cos^2 \theta} = 2 \int \frac{\sin \theta \, d\theta}{\sin^2 \theta \cos^2 \theta} = 2 \int \frac{\sin \theta \, d\theta}{(1 - \cos^2 \theta) \cos^2 \theta}.$$

Answer 3

$$\int \frac{\sqrt{x^2+4}} x$$ $$=\int \frac{2\sec^2\theta\sqrt{4\tan^2\theta+4}}{2\tan \theta}$$ $$=2\int \frac{\sec^3\theta}{\tan \theta}$$ $$=2\int \frac{1/\cos^3\theta}{\sin \theta/\cos \theta}$$ $$=2\int \frac{1}{\sin \theta \cos^2 \theta}$$ $$=2\int \frac{\sin \theta }{\sin^2 \theta \cos^2 \theta}$$ $$=2 \int \frac{\sin \theta }{(1 - \cos^2 \theta) \cos^2 \theta}$$ $$=2 \int \frac{-1}{(1 - u^2 ) u^2 }$$ $$=2 \int \frac{-1}{u^2 - u^4 }$$ Rest should be easy....

If you want a different way to do this problem, try $x=2\sinh \theta$

If you want to do this another way, try Euler Substitutions!

Answer 4

Alternately, let $x=2\sinh t$, and use the fact that $\cosh^2t-\sinh^2t=1,~\sinh't=\cosh t$, and $\cosh't=\sinh t$.

Answer 5

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\overbrace{\color{#66f}{\large\int\frac{\root{x^{2} + 4}}{x}\,\dd x}} ^{\ds{x=t^{1/2}\ \dsc{\imp}\ t=x^{2}}}\ =\ \overbrace{\half\int\frac{\root{t + 4}}{t}\,\dd t}^{\ds{\dsc{t}=\dsc{\xi^{2} - 4}\ \imp\ \dsc{\xi}=\dsc{\root{t + 4}}}}\ =\ \int\frac{\xi^{2}}{\xi^{2} - 4}\,\dd\xi \\[5mm]&=\int\pars{1 + \frac{1}{\xi - 2} - \frac{1}{\xi + 2}}\,\dd\xi =\xi + \ln\pars{\verts{\frac{\xi - 2}{\xi + 2}}} =\root{t + 4} + \ln\pars{\verts{\frac{\root{t + 4} - 2}{\root{t + 4} + 2}}} \\[5mm]&=\color{#66f}{\large\root{x^{2} + 4} +\ln\pars{\verts{\frac{\root{x^{2} + 4} - 2}{\root{x^{2} + 4} + 2}}}} + \mbox{a constant} \end{align}