I have a problem where i am not sure how the solutions manual got the step. I understand they are taking
integration by parts twice but i dont see how they are producing the 6 in the second integration.
∫x3exdx
u=x3,dv=ex
Take integration by parts 3 times..
u=x,v=ex,du=dx
Now, lets put it in the form ∫udv=uv−∫vdu
We have
∫x3exdx=x3ex−3∫x2exdx
Now from here we need to do integration by parts on the integral again because it doesnt fit any rules?
But when i do
u=x2,dv=ex
So, then i would have du=2xdx,v=ex
now, this is where i am not sure how they get 6?
=x3ex−3x2ex+6∫xexdx
Do i need to distribute the -3 to the integral and times it by the 2xdx?
integration by parts twice but i dont see how they are producing the 6 in the second integration.
∫x3exdx
u=x3,dv=ex
Take integration by parts 3 times..
u=x,v=ex,du=dx
Now, lets put it in the form ∫udv=uv−∫vdu
We have
∫x3exdx=x3ex−3∫x2exdx
Now from here we need to do integration by parts on the integral again because it doesnt fit any rules?
But when i do
u=x2,dv=ex
So, then i would have du=2xdx,v=ex
now, this is where i am not sure how they get 6?
=x3ex−3x2ex+6∫xexdx
Do i need to distribute the -3 to the integral and times it by the 2xdx?