Question

Evaluate the following integrals.
$\int \frac{dx}{\sqrt{2x-3x^2+1}}$

Answer 1

$2x-3x^2+1=-3x^2+2x+1$

$⟹\frac{4}{3}-(\sqrt{3}x-\frac{1}{\sqrt{3}}{)}^2$

$\int \frac{dx}{\sqrt{\frac{4}{3}-(\sqrt{3}x-\frac{1}{\sqrt{3}}{)}^2}}$

Putting $\frac{3x-1}{2}=t$

$⟹\frac{3}{2}dx=dt$

$⟹\int \frac{2dt}{\sqrt{3}\sqrt{4-4t^2}}$

$⟹\frac{1}{\sqrt{3}}\int \frac{dt}{\sqrt{1-t^2}}$

$⟹\frac{si{n}^{-1}\frac{3x-1}{2}}{\sqrt{3}}+C$