$\int \frac{x}{\sqrt{x^2+2}} \, \operatorname d\!x$ via substitution How would you do this using u-substitution? 
$$
\int \frac{x}{\sqrt{x^2+2}} \,  \operatorname d\!x
$$
 A: Let $u=x^2+2$ so that $du =2x\,dx$. That should get you started.
A: $$
\int \frac{\bf{x}}{\sqrt{x^2+2}} \, \bf{dx}
$$
Let $u=x^2+2\,;\;$ then $\;du =2x\,dx \implies\; \color{blue}{\bf{\frac 12 \,du}} \,=\, \bf{x\,dx}\,;\;$ this gives us
$$
\int \color{blue}{\bf{\frac 12}} \frac{\color{blue}{\bf{du}}}{\sqrt{u}} \, = \,\frac 12 \int\; u^{-1/2}\,du
$$ 
$$\,\frac 12 \int\; u^{-1/2}\,du = \frac 12 \left(\frac{u^{1/2}}{\large\frac 12} + C \right)\;=\; \sqrt u + C = \sqrt{x^2 + 2} + C$$
Just remember after integrating in terms of $u$, to replace $u$ with $x^2 + 2$.
A: $u=x^2+2\Rightarrow du=2xdx\Rightarrow xdx=\frac{1}{2}du$
A: Take $u=\sqrt{x^2+2}$. Then $du=\cdots$. 
A: The "hard" part is that there is a bunch of stuff under a square root sign. So you make a substitution to simplify that. The simplest way is $u = x^2+2$. Other options are $x^2 = 2 \tan^2 u$ and $x^2 = 2 \sinh^2 u$ (so as to invoke the pythagorean identities to simplify the square root), but in this case they are more complicated than you need.
A: Directly: if $\,f(x)\,$ is a derivable function, then
$$\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+C$$
In our case
$$\frac{d}{dx}\left(x^2+2\right)=2x\Longrightarrow \frac{1}{2}\int\frac{2x}{\sqrt{x^2+2}}dx=\sqrt{x^2+2}+C$$
A: Keep in mind, in order for your substitution to be successful it is usually the case that you must see a multiple of the derivative of the thing you're substituting as a factor in your integrand. Observe that the $x^2$ term under the radical has as its derivative $2x$, and $x$ is a factor of the integrand. So any substitution involving $x^2$ has a chance of working. 
