Integrate the function: $\frac{4x+1}{\sqrt{2x^{2}+x-3}}$

(N/A) Let $2x^{2}+x-3 = t$.
Then,differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(2x^{2}+x-3) = \frac{dt}{dx}$
$(4x+1) dx = dt$.
Substituting these into the integral:
$\int \frac{4x+1}{\sqrt{2x^{2}+x-3}} dx = \int \frac{1}{\sqrt{t}} dt$.
This is equivalent to $\int t^{-1/2} dt$.
Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:
$= \frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C$.
Substituting the value of $t$ back:
$= 2\sqrt{2x^{2}+x-3} + C$,where $C$ is an arbitrary constant.