Помогите решить 2 и 3! СРОЧНО!!!!!

Question 1

Question 2

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

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