Question

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

Answer 1

Let $4x+1=A\frac{d}{dx}(2x^2+x-3)+B$
$\Rightarrow 4x+1=A(4x+1)+B$
$\Rightarrow 4x+1=4Ax+A+B$
Equating the coefficients of $x$ and constant term on both sides, we obtain
$4A=4\Rightarrow A=1$
$A+B=1\Rightarrow B=0$
Let $2x^2+x-3=t$
$\therefore (4x+1)dx=dt$
$\Rightarrow \int \frac{4x+1}{\sqrt{2x^2+x-3}}dx=\int \frac{1}{\sqrt{t}}dt$
$=2\sqrt{t}+C=2\sqrt{2x^2+x-3}+C$