Prove

Let $4 x+1=A \frac{d}{d x}\left(2 x^{2}+x-3\right)+B$

$\Rightarrow 4 x+1=A(4 x+1)+B$

$\Rightarrow 4 x+1=4 A x+A+B$

Equating the coefficients of x and constant term on both sides, we obtain

$4 A=4 \Rightarrow A=1$

$A+B=1 \Rightarrow B=0$

Let $2 x^{2}+x-3=t$

$\therefore(4 x+1) d x=d t$

$\Rightarrow \int \frac{4 x+1}{\sqrt{2 x^{2}+x-3}} d x=\int \frac{1}{\sqrt{t}} d t$

$=2 \sqrt{t}+\mathrm{C}$

$=2 \sqrt{2 x^{2}+x-3}+\mathrm{C}$