I need to solve
$$\int\frac{1}{\sqrt{x^2+1}}$$
but I'm nowhere near the solution. I tried substituting
$u = x^2 + 1$
such that $dx = \frac{1}{2x}$ yielding $$\int u^{-1/2} du = \frac{1}{2x}\frac{u^{1/2}}{1/2}=\frac{\sqrt{u}}{x} =\frac{\sqrt{x^2+1}}{x}$$
but this is nothing near the solution of
$$\int\frac{1}{\sqrt{x^2+1}} = ln|\sqrt{x^2+1}+x|+C$$
Hint: Substituting $$x=\tan(t)$$ then we get $$x^2+1=\tan^2(t)+1=\frac{\sin^2(t)+\cos^2(t)}{\cos^2(t)}=\frac{1}{\cos^2(t)}$$ and $$dx=(\tan^2(t)+1)dt$$
Use the trigonometric substitution $x(\theta)=\tan{\theta}$:
$$\int\frac{1}{\sqrt{x^2+1}}\,dx= \int\frac{1}{\sqrt{[x(\theta)]^2+1}}x'(\theta)\,d\theta=\\ \int\frac{1}{\sqrt{\tan^2{\theta}+1}}\sec^2{\theta}\,d\theta= \int\frac{\sec^2{\theta}}{\sqrt{\sec^2{\theta}}}\,d\theta=\\ \int\frac{\sec^2{\theta}}{|\sec{\theta}|}\,d\theta. $$
Here, $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$ because on that interval $\tan\theta$ gives you the entire real line and $\sec{\theta}\ge 1$ there. Therefore, $|\sec\theta|=\sec\theta$:
$$ \int\frac{\sec^2{\theta}}{|\sec{\theta}|}\,d\theta= \int\frac{\sec^2{\theta}}{\sec{\theta}}\,d\theta= \int\sec{\theta}\,d\theta. $$
Now just use the following fact: $$\int\sec{x}\,dx=\ln{|\tan{x}+\sec{x}|}+C.$$
