Question

Solve ${\int }^{}\frac{1}{\sqrt{2x^2+3x+5}}dx$

Answer 1

$\int \frac{dx}{\sqrt{2x^2+3x+5}}$

Lets take $Y=2x^2+3x+5$

$Y={(\sqrt{2}x+\frac{3}{2\sqrt{2}})}^2+5-\frac{9}{8}$

$Y={(\sqrt{2}x+\frac{3}{2\sqrt{2}})}^2+\frac{31}{8}$

$\int \frac{dx}{\sqrt{2x^2+3x+5}}=\int \frac{dx}{{(\sqrt{2}x+\frac{3}{2\sqrt{2}})}^2+{\left(\sqrt{\frac{31}{8}}\right)}^2}$

Put $\sqrt{2}x+\frac{3}{2\sqrt{2}}=t$ .$\sqrt{2}dx=dt$

$=\frac{1}{\sqrt{2}}\int \frac{dt}{\sqrt{t^2+{\left(\frac{\sqrt{31}}{\sqrt{8}}\right)}^2}}$

$=\frac{2\sqrt{2}}{\sqrt{2}\sqrt{31}}ta{n}^{-1}\left(\frac{t}{\left(\frac{\sqrt{31}}{\sqrt{8}}\right)}\right)+c$

$=\frac{2}{\sqrt{31}}ta{n}^{-1}\left(\frac{4x+3}{\sqrt{31}}\right)+c$