Question

solution on textbook says
Let dv = (1+2x)^-2 dx
-> v = - 2*(2x+1)^-1
I know - 2*(2x+1)^-1 is the integral of
(1+2x)^-2 dx
but I let dv equal (1+2x)^-2 dx
and I don't know where's d been(I mean d in front of dv)

Answer 1

what's the integral of dv?

\[\int \mathrm dv=?\]

Answer 2

There's always something in fornt of dv
in my opinion

Answer 3

\[\mathrm dv = (1+2x)^{-2} \mathrm dx\\\int 1\mathrm dv = \int (1+2x)^{-2} \mathrm dx\]
just integrate both sides

Answer 4

∫(xe^2x)/(1+2x)^2 dx

Answer 5

I know I should use intergral by parts

Answer 6

and I can solve it but solution on textbook confused me alot

Answer 7

it saids let u = e(x)^2 , dv = 1/(1+2x)^2 dx -> du = (x*(e)^2x + e^2x) dx

Answer 8

and v = - 1 / 2(1+2x)
I'd like to know why du is used not in a full sentence but in a strange one

Answer 9

sorry, what would you like to integrate

Answer 10

I'm working on this question
∫(xe^2x)/(1+2x)^2 dx

Answer 11

ok thanks

Answer 12

I can solve this problem
but I don't really understand the solutions on the textbook

Answer 13

\[\Large \int \left[x e^{2x}\right] \left[\frac{1}{(1+2x)^2}\mathrm dx\right]\]
the trick in the integration by parts is to separate the integrand into two parts \(u\) that can be differentiated, and \(\mathrm dv\) that can be integrated.

Answer 14

UH I know that but I don't know what can a single dv do

Answer 15

if your wondering about where the dv goes in the eq. well we just dont use it but we use it in the set up process, i am not really sure as to what your trying to get to, please explin further