[m] ∫ \frac{du}{u^5}=\frac{u^{-5+1}}{-5+1}+C[/m]
u=1-3x
du=(1-3x)`dx=-3dx
dx=(-1/3)du
[m] ∫ \frac{4dx}{(1-3x)^5}=∫ \frac{4\cdot (-\frac{1}{3})du}{u^5}=[/m]
[m]=4\cdot (-\frac{1}{3}∫ \frac{du}{u^5}=4\cdot (-\frac{1}{3})\cdot \frac{u^{-5+1}}{-5+1}+C=[/m]
[m]=4\cdot (-\frac{1}{3})\cdot \frac{(1-3x)^{-5+1}}{-5+1}+C=\frac{1}{3(1-3x)^4}+C[/m]
6)
[m]∫ sinu du=-cosu+C[/m]
u=2x
du=2dx
dx=(1/2)du
[m]∫ sin2x dx=∫ sinu (\frac{1}{2})du=\frac{1}{2}\cdot (-cosu)+C=-\frac{1}{2}\cdot cos2x+C[/m]