Задача 43663 ...

sova

2020-01-25 10:07:11

Замена
[m]\frac{1}{x}=t[/m]

⇒

[m]x=\frac{1}{t}[/m]

[m]dx=-\frac{1}{t^2}[/m]


[m]\int \frac{dx}{x^2\sqrt{3+x^2}}=\int \frac{(-\frac{dt}{t^2})}{\frac{1}{t^2}\sqrt{3+\frac{1}{t^2}}}=-\int \frac{tdt}{\sqrt{3t^2+1}}=-\frac{1}{2\sqrt{3}}\int \frac{d(t^2+\frac{1}{3})}{\sqrt{t^2+\frac{1}{3}}}[/m]


[m]-\frac{1}{2\sqrt{3}}\int (t^2+\frac{1}{3})^{-\frac{1}{2}} d(t^2+\frac{1}{3})=-\frac{1}{2\sqrt{3}}\frac{(t^2+\frac{1}{3})^{-\frac{1}{2}+1}}{(-\frac{1}{2}+1)}+C=-\frac{1}{\sqrt{3}}\sqrt{t^2+\frac{1}{3}}+C[/m]


[m]t=\frac{1}{x}[/m]