Question

Помогите решить пожалуйста!!

Answer 1

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

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

$=frac{pi}{4}-ln2-frac{sqrt3}{3}cdot frac{pi}{6}+lnfrac{4}{3}=frac{pi}{4}-ln2-frac{sqrt3pi }{18}+2ln2-ln3=\\=frac{pi}{4}-frac{sqrt3pi }{18}+ln2-ln3=frac{(9-2sqrt3)pi }{36}+lnfrac{2}{3}; .$