Evaluating the integral $ \int{\frac{x}{\sqrt{2x^2 + 3}}}dx $

Question

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?

Answer 1

$$\int\frac{x}{\sqrt{2x^2+3}}dx=\frac{1}{2}\int\underbrace{\frac{4x}{2\sqrt{2x^2+3}}}_{\text{of the form} \ \frac{u'}{2\sqrt u}}dx=\frac{\sqrt{2x^2+3}}{2}+C$$

Answer 2

The square plus constant beneath the square root in the denominator definitely suggests that a trig substitution is helpful! However, if we notice that the numerator is almost the derivative of what's under the denominator, the problem falls apart. Set $$u = 2x^2+3.$$

Answer 3

$$ \int \frac{x}{\sqrt{2x^2+3}}dx $$ 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 $$ \frac{1}{2}\int du= \frac{1}{2}u+C = \frac{1}{2}\sqrt{2x^2+3}+C $$