1. интеграл dx/2+3x^2 2. интеграл 6z^2 da/(1-2z^3)^4 пж

Решите задачу:

$1)\quad \int \frac{dx}{2+3x^2} =\int \frac{dx}{3(\frac{2}{3}+x^2)}=\frac{1}{3}\cdot \frac{1}{\sqrt{2/3}}\cdot arctg \frac{x}{\sqrt{2/3}} +C=\\\\= \frac{1}{\sqrt6} \cdot arctg \frac{\sqrt3\, x}{\sqrt2} +C\\\\2)\quad \int \frac{6z^2\, dz}{(1-2z^3)^4}=\\\\=[\, t=1-2z^3\; ,\; dt=-6z^2\, dz\; ,\; dz=-\frac{dt}{6z^2} \; ,\; 6z^2\, dz=-dt\, ]=\\\\=-\int \frac{dt}{t^4} =-\int t^{-4}dt=-\frac{t^{-3}}{-3}+C=\frac{1}{3t^3}+C=\frac{1}{3(1-2z^)^3}+C$