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Going further with integration by parts. Don't know whether to proceed further or not 
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#1
Jan2813, 08:13 PM

P: 34

Hey guys,
Need you push to proceed further with integration by parts: ∫e^{3x}*3*x^{2}*ydx=y∫e^{3x}*3*x^{2}dx setting u=3*x^{2}du=6*x dx dv= e^{3*x}dx v= 1/3* e^{3*x} ∫ e^{3*x}*3*x^{2}*ydx=y*(3*x^{2}* 1/3* e^{3*x}∫6*x*1/3* e^{3*x}dx) =y*(3*x^{2}* 1/3* e^{3*x}6/3*∫x*e^{3*x}dx) Solving further about x*e^{3*x} u=xdu=dx dv=e^{3*x}dxv=1/3*e^{3*x} ∫ e^{3*x}*3*x^{2}*ydx=y*(3*x^{2}* 1/3* e^{3*x}6/3*(x*1/3*e^{3*x}∫1/3e^{3*x}) 


#2
Jan2813, 08:14 PM

P: 34

How can we go further with solution as exp(3*x) repeats all the time?



#3
Jan2813, 08:53 PM

Mentor
P: 21,216

You've done all the hard work. ∫e^{3x}dx is easy, using a simple substitution.



#4
Jan2913, 08:50 AM

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P: 39,338

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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