Question

Плиз помогите срочно

Answer 1

1) $intlimits^1_0 { frac{4xdx}{ sqrt[3]{(9x-1)^2} 1 sqrt[3]{9x-1} +1} }=I$
Замена
$sqrt[3]{9x-1} =y; x= frac{y^3+1}{9} ; dx= frac{3y^2}{9} dy= frac{y^2}{3} dy; y(0)=-1;y(1)=2$
$I= intlimits^2_{-1} { frac{4(y^3+1)/9}{y^2-y+1} } , frac{y^2}{3}dy= frac{4}{27} intlimits^2_{-1} { frac{(y+1)(y^2-y+1)*y^2}{y^2-y+1} }, dy= frac{4}{27} intlimits^2_{-1} {y^2(y+1)} , dy=$
$= frac{4}{27} intlimits^2_{-1} {(y^3+y)} , dy= frac{4}{27}( frac{y^4}{4}+ frac{y^2}{2} )| _{-1} ^{2}= frac{4}{27}( frac{16}{4}+ frac{4}{2}- frac{1}{4}- frac{1}{2} )=$
$= frac{4}{27}* frac{21}{4}= frac{7}{9}$

2) $intlimits^1_0 {x^2arcctgx} , dx =I$
Берем по частям. $u = arcctg x; dv = x^2 dx; du = - frac{dx}{1+x^2}; v = frac{x^3}{3}$
$I=u*v- intlimits^1_0 {v} , du= frac{x^3}{3}*arcctgx- intlimits^1_0 { frac{-x^3}{3(1+x^2)} } , dx=$
$=frac{x^3}{3}*arcctgx+ frac{1}{3} intlimits^1_0 { frac{x^3}{1+x^2} } , dx =frac{x^3}{3}*arcctgx+ frac{1}{3} intlimits^1_0 { frac{x^3+x-x}{x^2+1} } , dx =$
$=frac{x^3}{3}*arcctgx+ frac{1}{3} intlimits^1_0 {(x- frac{x}{x^2+1} )} , dx =frac{x^3}{3}*arcctgx+ frac{1}{3} ( frac{x^2}{2} - intlimits^1_0 { frac{x}{x^2+1} } , dx )$
$=frac{x^3}{3}*arcctgx+ frac{x^2}{6}- frac{1}{6} intlimits^1_0 { frac{2xdx}{x^2+1} } = frac{x^3}{3}*arcctgx+ frac{x^2}{6}- frac{1}{6} intlimits^1_0 { frac{d(x^2+1)}{x^2+1} } =$
$=(frac{x^3}{3}*arcctgx+ frac{x^2}{6}- frac{1}{6} ln|x^2+1| )|_{0}^{1} = frac{1}{3}frac{ pi }{4}+ frac{1}{6}- frac{1}{6}ln2-0+frac{1}{6} ln1=$
$= frac{ pi }{12}+ frac{1}{6}- frac{ln2}{6}=frac{ pi+2-2ln2 }{12} = frac{ pi +2-ln4}{12}$