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I would like to ask for some help regarding the following indefinite integral, tried integration by parts and trigonometric substitution which both brought me to $\int\frac{\sec\theta}{\tan\theta}d\theta$, and from this point it is messy to integrate by parts, any help would be appreciated.

$$\int\frac{dx}{x\sqrt{x^2+1}}$$

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    $\begingroup$ Notice that $\frac{\sec \theta}{\tan \theta} = \csc \theta$. $\endgroup$ – Travis Sep 5 '14 at 14:46
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    $\begingroup$ Oh well, this substitution didn't cross my mind to be honest, I kinda feel stupid, thank you. $\endgroup$ – user122673 Sep 5 '14 at 14:49
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    $\begingroup$ You're welcome. When in doubt, you can always write such expressions in terms of, say, just sine and cosine, and cancel from there. $\endgroup$ – Travis Sep 5 '14 at 15:26
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$$\int \frac{\sec \theta}{\tan\theta} = \frac{\frac 1{\cos \theta}}{\frac {\sin\theta}{\cos\theta}}\,d\theta = \int \frac 1{\sin\theta}\,d\theta = \int \csc\theta \,d\theta$$

Alternatively, given $$\int\frac{dx}{x\sqrt{x^2+1}} = \int\frac{x\,dx}{x^2 \sqrt{x^2 + 1}}$$

$$\text{Put }\;x^2 + 1 = u^2\;\iff \;x^2 = u^2 - 1\; \implies \;u\,du = x\,dx$$ This gives us the integral, after substitution: $$\int \frac{u\,du}{(u^2-1)u}=\int \frac{du}{(u^2-1)} = \frac 12\int \left(\frac 1{u-1} - \frac 1{u+1}\right)\,du$$

I'm sure you can take it from here.

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Multiply the integrand by $\dfrac{x}{x}$, we will have $$ \int\frac{x\ dx}{x^2\sqrt{x^2+1}}\ dx. $$ Now, set $u^2=x^2+1\ \Rightarrow\ u\ du=x\ dx$ then \begin{align} \int\frac{x\ dx}{x^2\sqrt{x^2+1}}\ dx&=\int\frac{1}{u^2-1}\ du\\ &=\frac12\int\left[\frac1{u-1}-\frac1{u+1}\right]\ du. \end{align} The rest should be easy.

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Even another aproach to the integral (albeit rather in "philosophical" form in some sence); by substitution $x=1/t$ :

$$\int\frac{\mathrm{d}x}{x\sqrt{1+x^2}}=-\int\frac{\mathrm{d}t}{\sqrt{1+t^2}} = -\operatorname{arcsinh}t = -\operatorname{arcsinh}\frac{1}{x} = -\operatorname{arccsch}x + C $$

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