Question

Ydy/(3y^2+1)^(1/2) найти интеграл !

Answer 1

$\int \frac{y\, dy}{\sqrt{3y^2+1}}=\frac{1}{6}\int \frac{6y\, dy}{\sqrt{3y^2+1}}=\frac{1}{6}\int \frac{d(3y^2+1)}{\sqrt{3y^2+1}}=\\\\=[\, \int \frac{dt}{\sqrt{t}}=\int t^{\frac{-1}{2}}dt=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}}=2t^{\frac{1}{2}}+C=2\sqrt{t}+C\, ]=\\\\=\frac{1}{6}\cdot 2\sqrt{3y^2+1}+C$