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Question 1

Question 2

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

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