табличный интеграл [m] ∫ \frac{du}{u^2+a}=ln|u+\sqrt{u^2+a}|+C[/m] ( "длинный логарифм"), [m] u^2=3x^2; u=\sqrt{3}x[/m]
[m]du=\sqrt{3}dx[/m]
[m]dx=\frac{du}{\sqrt{3}}du=\frac{d(\sqrt{3}x)}{\sqrt{3}}[/m]
[m] ∫ \frac{dx}{\sqrt{3x^2+2}}= \frac{1}{\sqrt{3}}∫ \frac{d(\sqrt{3}x)}{\sqrt{(\sqrt{3}x)^2+2}}=\frac{1}{\sqrt{3}}ln|\sqrt{3}x+\sqrt{3x^2+2}|+C[/m]
Проверка
Так как
[m] |x|`=\left\{\begin {matrix}-1, x<0\\1, x>0\end {matrix}\right.[/m]
то рассматриваем случай
[m]\sqrt{3}x+\sqrt{2x^2+2} >0 ⇒ |\sqrt{3}x+\sqrt{3x^2+2}|=\sqrt{3}x+\sqrt{3x^2+2}[/m]
тогда
[m] (\frac{1}{\sqrt{3}}ln(\sqrt{3}x+\sqrt{3x^2+2})+C)`=\frac{1}{\sqrt{3}}\cdot (ln(\sqrt{3}x+\sqrt{3x^2+2}) )`+C`=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot ((\sqrt{3}x+\sqrt{3x^2+2})`+0=[/m]
[m]=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot ((\sqrt{3}x)`+(\sqrt{3x^2+2})`)=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot (\sqrt{3}+\frac{1}{2\sqrt{3x^2+2}}\cdot (3x^2+2)`)=[/m]
[m]=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot (\sqrt{3}+\frac{1}{2\sqrt{3x^2+2}}\cdot 6x=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot (\sqrt{3}+\frac{3x}{\sqrt{3x^2+2}})=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot \frac{(\sqrt{3}\cdot (\sqrt{3x^2+2}+\sqrt{3}x)}{\sqrt{3x^2+2}})=[/m]
[m]=\frac{1}{\sqrt{3x^2+2}}[/m]
Если
[m]\sqrt{3}x+\sqrt{2x^2+2} <0 ⇒ |\sqrt{3}x+\sqrt{3x^2+2}|=-\sqrt{3}x-\sqrt{3x^2+2}[/m]
тогда
[m] (\frac{1}{\sqrt{3}}ln(-\sqrt{3}x-\sqrt{3x^2+2}))+C)`=\frac{1}{\sqrt{3}}\cdot (ln(-\sqrt{3}x-\sqrt{3x^2+2})) )`+C`=\frac{1}{\sqrt{3}}\cdot\frac{1}{(-\sqrt{3}x-\sqrt{3x^2+2})}\cdot (-\sqrt{3}x-\sqrt{3x^2+2})`+0=[/m]
далее аналогично
[m]=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot ((\sqrt{3}x)`+(\sqrt{3x^2+2})`)=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot (\sqrt{3}+\frac{1}{2\sqrt{3x^2+2}}\cdot (3x^2+2)`)=[/m]
[m]=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot (\sqrt{3}+\frac{1}{2\sqrt{3x^2+2}}\cdot 6x=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot (\sqrt{3}+\frac{3x}{\sqrt{3x^2+2}})=\frac{1}{\sqrt{3}}\cdot\frac{1}{(\sqrt{3}x+\sqrt{3x^2+2})}\cdot \frac{(\sqrt{3}\cdot (\sqrt{3x^2+2}+\sqrt{3}x)}{\sqrt{3x^2+2}})=[/m]
[m]=\frac{1}{\sqrt{3x^2+2}}[/m]