Решить интеграл 1) dx/3x^2+8x-2 2)(2-cosx/3)dx

Решите задачу:

$1)\; \int \frac{dx}{3x^2+8x-2}=\Big [\, 3x^2+8x-2=3(x^2+\frac{8}{3}x-\frac{2}{3})=\\\\=3\cdot \Big ((x+\frac{8}{6})^2-\frac{16}{9}-\frac{2}{3}\Big )=3\cdot \Big ((x+\frac{4}{3})^2-\frac{22}{9}\Big )\, \Big ]=\\\\=\frac{1}{3}\int \frac{dx}{(x+\frac{4}{3})^2-\frac{22}{9}}=[\, t=x+\frac{4}{3}\; ,\; dt=dx\, ]=\frac{1}{3}\int \frac{dt}{t^2-\frac{22}{9}}=\\\\=\frac{1}{3}\cdot \frac{3}{\sqrt{22}}\cdot ln\Big |\frac{t-\frac{\sqrt{22}}{3}}{t+\frac{\sqrt{22}}{3}}\Big |+C=\frac{1}{\sqrt{22}}\cdot ln\Big |\frac{3x+4-\sqrt{22}}{3x+4+\sqrt{22}}\Big |+C$

$2)\; \; \int (2-cos\frac{x}{3})dx=2\int dx-3\int \frac{1}{3}\cdot cos\frac{x}{3}\, dx=\\\\=2x-3\int cos\frac{x}{3}\cdot d( \frac{x}{3})=2x-3\cdot sin \frac{x}{3}+C\\\\\star \; [\; d(\frac{x}{3})=\frac{1}{3}dx\; ]$