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Going further with integration by parts. Don't know whether to proceed further or not

by Ein Krieger
Tags: integration, parts, proceed
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Ein Krieger
#1
Jan28-13, 08:13 PM
P: 34
Hey guys,

Need you push to proceed further with integration by parts:

∫e3x*3*x2*ydx=y∫e3x*3*x2dx

setting u=3*x2-------du=6*x dx
dv= e3*xdx--- v= 1/3* e3*x

∫ e3*x*3*x2*ydx=y*(3*x2* 1/3* e3*x-∫6*x*1/3* e3*xdx)
=y*(3*x2* 1/3* e3*x-6/3*∫x*e3*xdx)
Solving further about x*e3*x
u=x---du=dx
dv=e3*xdx---v=1/3*e3*x
∫ e3*x*3*x2*ydx=y*(3*x2* 1/3* e3*x-6/3*(x*1/3*e3*x-∫1/3e3*x)
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Ein Krieger
#2
Jan28-13, 08:14 PM
P: 34
How can we go further with solution as exp(3*x) repeats all the time?
Mark44
#3
Jan28-13, 08:53 PM
Mentor
P: 21,313
You've done all the hard work. ∫e3xdx is easy, using a simple substitution.

HallsofIvy
#4
Jan29-13, 08:50 AM
Math
Emeritus
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Thanks
PF Gold
P: 39,568
Going further with integration by parts. Don't know whether to proceed further or not

If you have [itex]\int x^n f(x)dx[/itex], where "f" is easy to integrate any number of times (and the "nth" integral of [itex]e^{3x}[/itex] is [itex](1/3^n)e^{3x}[/itex]), just continue taking [itex]u= x^n[/itex], [itex]dv= f(x)dx[/itex]. Everytime du will have x to a lower power until, eventually, it is just [itex]x^0= 1[/itex].


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