Помогите решить 2 и 3! СРОЧНО!!!!!

2) $intlimits { frac{x^2}{(4x^3+3)^3}} , dx;t=4x^3;dt=12x^32dx;$
$intlimits { frac{x^2}{(4x^3+3)^3}}, dx= frac{1}{12} intlimits {(t+3)^{-3}} , dt=- frac{1}{24(t+3)^{2}}+C=$
$=- frac{1}{24(4x^3+3)^2}+C=$
$intlimits {x^4lnx} , dx;$
$u=lnx;$ $du= frac{dx}{x;}$
$dv=x^4dx;$ $v= frac{x^5}{5};$
$intlimits {x^4lnx} , dx= frac{x^5}{5}lnx- frac{1}{5}intlimits {x^4} , dx = frac{1}{5}x^5lnx- frac{1}{25}x^5+C;$
$intlimits { frac{3x^2+8}{x^3+4x^2+4}} , dx= intlimits { frac{3x^2+8}{x(x^2+4x+4)}}, dx= intlimits { frac{3x^2+8}{x(x+2)^2}}, dx$
$frac{3x^2+8}{x(x+2)^2}= frac{A}{x} + frac{B}{x+2} + frac{C}{(x+2)^2}= frac{A(x+2)^2+Bx(x+2)+Cx}{x(x+2)^2}=$
$= frac{A x^{2} +4Ax+4A+Bx^2+2Bx+Cx}{x(x+2)^2}= frac{(A+B) x^{2} +(4Ax++2B+C)x+4A}{x(x+2)^2};$
$left { {{B=1}atop {4Ax+2B+C=0}}atop {4A=8}} right.; left { {{A+B=3}atop {C=-10}}atop {A=2}} right$
$frac{3x^2+8}{x(x+2)^2}= frac{2}{x} + frac{1}{x+2} - frac{10}{(x+2)^2}$
$intlimits { frac{3x^2+8}{x^3+4x^2+4}} , dx= intlimits { frac{3x^2+8}{x(x+2)^2}}, dx= intlimits { frac{2}{x}}, dx+ intlimits { frac{1}{x+2}}, dx- intlimits { frac{10}{(x+2)^2}}, dx=$
$2 intlimits { frac{1}{x}}, dx+ intlimits { frac{1}{x+2}}, dx-10 intlimits { frac{1}{(x+2)^2}}, dx=2lnx+ln(x+2)+ frac{10}{x+2}+C;$
3) $y= sqrt[5]{(x+2)^2}* sqrt{x-4}*(x^2-1)^3;$

$y=frac{2}{5}(x+2)^{- frac{3}{5}}* sqrt{x-4}*(x^2-1)^3+$
$+sqrt[5]{(x+2)^2}*frac{1}{2sqrt{x-4}}*(x^2-1)^3+ sqrt[5]{(x+2)^2}* sqrt{x-4}* 3(x^2-1)^2*2x=$
$=frac{(x^2-1)^2}{ sqrt[]{(x+2)^3}sqrt{x-4}}(frac{2}{5}* (x-4)*(x^2-1)+$
$+frac{1}{2} (x+2)*(x^2-1)+6x (x+2)*(x-4));$
$y=cos(cosx);$
$y=-sin(cosx)*(-sinx)=sin(cosx)*sinx;$