Question

$I=\int \frac{1}{2x^2+3x+4}dx$

Answer 1

$\int \frac{1}{2x^2+3x+4}dx=\frac{1}{2}\int \frac{1}{x^2+\frac{3}{2}x+2}dx=\frac{1}{2}\int \frac{1}{{(x+\frac{3}{4})}^2+{\left(\frac{\sqrt{23}}{4}\right)}^2}dx=\frac{1}{2}\times \frac{4}{\sqrt{23}}{\tan }^{-1}\frac{x+\frac{3}{4}}{\frac{\sqrt{23}}{4}}+c=\frac{2}{\sqrt{23}}\times {\tan }^{-1}\frac{4x+3}{\sqrt{23}}+c$