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$F(x)=\int (cosx-\frac{1}{\sqrt{x^2+4}})dx=sinx-ln|x+\sqrt{x^2+4}|+C\\\\2)\; \; \int arctgx\, dx=[u=arctgx\; ,\; du=\frac{dx}{1+x^2}\; ,\; dv=dx\; ,\; v=x\; ]=\\\\==uv-\int v\, du=x\cdot arctgx-\int \frac{x\, dx}{1+x^2}=x\cdot arctgx-\frac{1}{2}\int \frac{2x\, dx}{1+x^2}=\\\\=[\, t=1+x^2,\; dt=2xdx\; ,\int \frac{dt}{t}=ln|t|+C\, ]=\\\\=x\cdot arctgx-ln|1+x^2|+C=x\cdot arctgx-ln(1+x^2)+C\; ;\\\\\int _{\frac{\sqrt3}{3}}^1arctgx\, dx=(x\cdot arctgx-ln(1+x^2))\, |_{\frac{\sqrt3}{3}}^1=$

$=arctg1-ln2-$ $(\frac{\sqrt3}{3}arctg\frac{\sqrt3}{3}-ln(1+\frac{1}{3}))=$

$=\frac{\pi}{4}-ln2-\frac{\sqrt3}{3}\cdot \frac{\pi}{6}+ln\frac{4}{3}=\frac{\pi}{4}-ln2-\frac{\sqrt3\pi }{18}+2ln2-ln3=\\\\=\frac{\pi}{4}-\frac{\sqrt3\pi }{18}+ln2-ln3=\frac{(9-2\sqrt3)\pi }{36}+ln\frac{2}{3}\; .$