Решите пожалуйста) а то не понятно как ((

1)∫dx/(4-3x)²= 4-3x=t, -3dx=dt =-1/3∫dt/t²=-1/3.(-1/t)=1/3t=1/3(4-3x)+c 2)∫∛(3x+1)²dx= 3x+1=t, 3dx=dt, dx=1/3dt =1/3∫∛t²dt=1/3.∫tˇ(2/3)dt=1/3tˇ(5/3)/(5/3)=(1/3)(3/5).∛tˇ5= =1/5.∛(3x+1)ˇ5+c 3)∫cotg(x/2)dx=2.∫cotgtdt=2.ln/sint/=2.ln/x/2/+c x/2=t, dx/2=dt, dx=2dt 4)-∫aˇ(xˇ4).x³dx= xˇ4=t,4x³dx=dt =-1/4∫aˇtdt=-1/4.aˇt/lna=-1/4.lna . aˇ(xˇ4)+c