calculus - Integrate $intfrac{dx}{xsqrt{x^2+1}}$

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.

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

Answer

$$\int \frac{\sec \theta}{\tan\theta} = \frac{\frac 1{\cos \theta}}{\frac {\sin\theta}{\cos\theta}}\,d\theta = \int \frac 1{\sin\theta}\,d\theta = \int \csc\theta \,d\theta$$

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.