Question

Вычислить определенный интеграл

Answer 1

$intlimits^0_{-8} , frac{dx}{5- sqrt[3]{x^2} } =[; x=t^3; ,; dx=3t^2, dt; ,; t_1=-2; ,; t_2=0; ]=\\= intlimits_{-2}^0 frac{3t^2, dt}{5-t^2} =-3cdot intlimits_{-2}^0frac{t^2, dt}{t^2-5} = -3cdot intlimits_{-2}^0Big (1+frac{5}{t^2-5} Big )dt=\\=-3cdot Big (t+5cdot frac{1}{2sqrt5}cdot lnBig | frac{t-sqrt5}{t+sqrt5} Big |Big )Big |_{-2}^0=\\=-3cdot Big ( frac{sqrt5}{2}cdot lnBig | frac{-sqrt5}{sqrt5} Big |+2-frac{sqrt5}{2}cdot lnBig |frac{-2-sqrt5}{-2+sqrt5}Big |Big )=$

$=-3cdot Big (frac{sqrt5}{2}cdot underbrace {ln1}_{0}+2- frac{sqrt5}{2}cdot lnBig | frac{-1}{9-4sqrt5} Big |Big )= frac{3sqrt5}{2}cdot lnfrac{1}{9-4sqrt5}-6; ;$

$2); ; intlimits^{11}_2 frac{sqrt{x-2}}{1+sqrt{x-2}}dx=[; x-2=t^2; ,; x=t^2+2; ,; dx=2t, dt; ,, \\t=sqrt{x-2}; ,; t_1=0; ,; t_2=sqrt{11-2}=3 ; ]= intlimits^3_0 frac{tcdot 2t, dt}{1+t}=\\=2cdot intlimits^3_0 Big (t-1+frac{1}{t+1}Big )dt=2 cdot Big ( frac{t^2}{2} -t+ln|t+1|)Big |_0^3=\\=2cdot (4,5-3+ln4)=3+2cdot ln4=3+2ln2^2=3+4cdot ln2$