Question

$I=\int \frac{1}{\sqrt{2x^2+3x+8}}dx$

Answer 1

$I=\int \frac{1}{\sqrt{2x^2+3x+8}}dx=\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{x^2+\frac{3}{2}x+4}}dx=\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{x^2+2\times \frac{3}{4}x+\frac{9}{16}+4-\frac{9}{16}}}dx=\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{{(x+\frac{3}{4})}^2+\frac{55}{16}}}dx=\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{{(x+\frac{3}{4})}^2+{\left(\frac{\sqrt{55}}{4}\right)}^2}}dx$

let $x+\frac{3}{4}=u$

$dx=du$

so $I=\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{u^2+{\left(\frac{\sqrt{55}}{4}\right)}^2}}du=\frac{1}{\sqrt{2}}\ln [u+\sqrt{u^2+{\left(\frac{\sqrt{55}}{4}\right)}^2}]+c$

putting $u=x+\frac{3}{4}$

$I=\frac{1}{\sqrt{2}}\ln [x+\frac{3}{4}+\sqrt{{(x+\frac{3}{4})}^2+{\left(\frac{\sqrt{55}}{4}\right)}^2}]+c=\frac{1}{\sqrt{2}}\ln [x+\frac{3}{4}+\sqrt{x^2+\frac{3}{2}x+\frac{9}{16}+\frac{55}{16}}]+c=\frac{1}{\sqrt{2}}\ln [x+\frac{3}{4}+\sqrt{x^2+\frac{3}{2}x+4}]+c$