<s>[INST] Evaluating the integral $ \int{\frac{x}{\sqrt{2x^2 + 3}}}dx $ I am trying to integrate the following:
$$
\int{\frac{x}{\sqrt{2x^2 + 3}}}dx
$$
It seems to me to be a trig  substitution; however, I couldn't seem to get it into one of the three forms, i.e., 
$$\sqrt{a^2 - x^2}$$
$$\sqrt{x^2 - a^2}$$
$$\sqrt{a^2 + x^2}$$
I also tried integration by parts. If I made $u = \frac{1}{\sqrt{2x^2 + 3}}$ and $dv = x$, the next integral was more complicated, and if I made $u = x$ and $dv = \frac{1}{\sqrt{2x^2 + 3}}$, I was again unsure how to integrate the 1/sqrt term.
How do I solve this?
 [/INST] $$ \int \frac{x}{\sqrt{2x^2+3}}dx $$
Let $u=\sqrt{2x^2+3}$, then
$$ \frac{d}{dx}u=\frac{2x}{\sqrt{2x^2+3}}\Rightarrow \frac{1}{2}du=\frac{x}{\sqrt{2x^2+3}}dx $$
So now we have
$$ \frac{1}{2}\int du= \frac{1}{2}u+C = \frac{1}{2}\sqrt{2x^2+3}+C $$
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