Question

xdx/(2x^2+x+5) неопределенный интеграл

Answer 1

$\displaystyle \int \dfrac{xdx}{2x^2+x+5}=\int\dfrac{4x+1}{4(2x^2+x+5)}dx-\int\dfrac{1}{4(2x^2+x+5)}dx=\\ \\ \\ =\dfrac{1}{4}\int\dfrac{(4x+1)dx}{2x^2+x+5}-\dfrac{1}{4}\int\dfrac{dx}{2x^2+x+5}=\dfrac{1}{4}\int\dfrac{d(2x^2+x+5)}{2x^2+x+5}-\\ \\ \\ -\dfrac{1}{4}\int\dfrac{dx}{(\sqrt{2}x+\frac{1}{2\sqrt{2}})^2+\frac{39}{8}}=\dfrac{1}{4}\ln\left|2x^2+x+5\right|-\dfrac{1}{4}\cdot \dfrac{2}{\sqrt{39}}{\rm arctg}\dfrac{4x+1}{\sqrt{39}}+C\\ \\ \\ =\dfrac{1}{4}\ln\left|2x^2+x+5\right|-\dfrac{1}{2\sqrt{39}}{\rm arctg}\dfrac{4x+1}{\sqrt{39}}+C$