Question

Solve
$\sqrt{2x^2+3x+4}dx$

Answer 1

$\to \int \sqrt{2(x^2+\frac{3}{2}n+2)}dx=\int \sqrt{2}\sqrt{{(x+\frac{3}{4})}^2}+2-\frac{9}{16}$

$=\int \sqrt{2}\sqrt{{(x+\frac{3}{4})}^2+\frac{23}{16}}=\int \sqrt{2}\sqrt{{(x+\frac{3}{4})}^2+{\left(\frac{\sqrt{23}}{4}\right)}^2}dx$

$=\sqrt{2}[\frac{x+3/4}{2}\sqrt{{(x+\frac{3}{4})}^2+{\left(\frac{\sqrt{23}}{4}\right)}^2}+\frac{23}{16\times 2}si{n}^{-1}\left(\frac{x+\frac{3}{4}}{\frac{\sqrt{23}}{4}}\right)]$

$=\sqrt{2}[\frac{4x+3}{8}\sqrt{x^2+\frac{3x}{2}+2}+\frac{23}{32}si{n}^{-1}\left(\frac{4x+3}{\sqrt{23}}\right)]+c$