Помогите решить интегралы при помощи подстановки, заранее благодарен

1)ʃdz/(5z+1)³u = 5z+1du/dz  = (5z+1)' = 5dz = du/5ʃdz/(5z+1)³ = ʃdu/5u³ =(1/5)ʃdu/u³ =-1/5*2u² =-1/10u² = -1/10(5z+1)² + c2)ʃdx/√(3x-1)³u = 3x-1du/dx  = (3x-1)' = 3dx = du/3ʃdx/√(3x-1)³ = (1/3)ʃdu/u^(3/2) = (1/3)(-2/√u) =-2/3√u = -2/3√(3x-1) + c3)ʃx³dx/(5x^4+3)⁵du/dx  = (5x^4+3)' = 20x³dx  = du/20x³ʃx³dx/(5x^4+3)⁵ = ʃx³du/(20x³)*u⁵ = (1/20)ʃdu/u⁵ =-(1/20)(1/4u^4) ==-1/80u^4 = -1/80(5x^4+3)^4 + c4)ʃcos(x)dx/√(1-sin(x))u = 1-sin(x)du/dx  = (1-sin(x))' = - cos(x)dx  = - du/cos(x)ʃcos(x)dx/√(1-sin(x)) = -ʃcos(x)du/cos(x)√u = -ʃdu/√u = -2√u = -2√(1-sin(x)) + c5)ʃx³dx/³√(5x^4 +2²)²u = 5x^4 +4du/dx  = (5x^4 +4)' = 20x³dx = du/20x³ʃx³dx/³√(5x^4 +2²)² = ʃx³du/20x³u^(2/3) = (1/20)ʃdu/u^(2/3) = (1/20)*3*³√u== (3*³√u)/20 = (3*³√(5x^4 +4))/20 + c