Cal
posted by renee .
USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL
ƪxe^2x dx

Please see your next post as there is the same problem.
Look at Related Questions below for they look similar?
Sra 
let u = x
du/dx = 1
du = dx
let dv = e^(2x)dx
dv/dx = e^(2x)
v = (1/2)e^(2x)
formula:
[int] u dv = uv  [int] v du
[int] x e^(2x) dx = (x)(1/2)e^(2x)  [int] (1/2)e^(2x) dx
= (1/2)xe^(2x)  (1/4)e^(2x)
Respond to this Question
Similar Questions

integration by parts
s integral s ln (2x+1)dx ? = ln(2x+1)x  s x d( ln (2x+1)) = ln(2x+1)x s x [(2x+1)'/ (2x+1)] dx = ln(2x+1)x s x [(2)/ (2x+1)] ? 
calc
how do you start this problem: integral of xe^(2x) There are two ways: 1) Integration by parts. 2) Differentiation w.r.t. a suitably chosen parameter. Lets do 1) first. This is the "standard method", but it is often more tedious than … 
calc
evaluate the integral: y lny dy i know it's integration by parts but i get confused once you have to do it the second time Leibnitz rule (a.k.a. product rule): d(fg) = f dg + g df y lny dy = d[y^2/2 ln(y)]  y/2 dy > Integral … 
Math  Indefinite Integration
Use integration by parts to solve: I = f (x^3)(e^(x^2)) dx (Let "f" represent the integral) Let u = (e^(x^2)), dx = (1) / (2x(e^(x^2))) du Let dv = x^3, v = (x^4) / (4) I = (1/4)(x^4)(e^(x^2))  (1/4) f (x^3) (2(e^(x^2))) du ... and … 
CALCULUS
USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪxe^2x dx 
CALCULUS
USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪx^5 Lnx dx 
calculus
USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪãx lnx dx 
cal
USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪsquare rootx lnx dx 
Maths  Integration by parts
Use the partial ractions to find The integral (3x^2x+2)/(x1)(x^2+1) dx 
Integration by Parts
integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=cos(t) i integral sin(t)e^(it)dt= e^(it)cos(t)+i*integral cost(t)e^(it)dt …