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

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.

HINT:

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

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

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

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

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