Question

anyone good with integration by parts?
Integrate ((x)e^(2x))/(1+2x)^2

Answer 1

\[\int\limits_{}^{}\frac{ xe ^{2x} }{ (1+2x)^2 }\]

Answer 2

ok ive gotten to where you are

Answer 3

i mean understanding what you did

Answer 4

small typo, let me fix it quick

Answer 5

Let :

\(u=xe^{2x} \implies du = (1\cdot e^{2x}+x\cdot 2e^{2x})dx = e^{2x}(1+2x)dx\)

and \(dv = \dfrac{1}{(1+2x)^2} dx\implies v = -\dfrac{1}{2(1+2x)}\)

Answer 6

\[\begin{align}\int\dfrac{ xe ^{2x} }{ (1+2x)^2 }~~ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} - \int e^{2x}(1+2x)\cdot \dfrac{-1}{2(1+2x)} dx\\~\\
&=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \int \dfrac{e^{2x}}{2}dx \\~\\
\end{align}\]

Answer 7

rest should be easy
is there any other function thats as easy as exponential function to integrate ?

Answer 8

so i should let u = e^2x and du = e^2

Answer 9

are you saying you dont know how to integrate \(e^{2x}\) ?

Answer 10

yea

Answer 11

u =2x, du =2, dv = e , v =e?

Answer 12

ohkie, its easy, whats the derivative of \(e^{2x}\) ?

Answer 13

e^(2x)?

Answer 14

heard of chain rule before ?

Answer 15

oh

Answer 16

e^(2x)2

Answer 17

right, so an antiderivative of e^(2x)2 is e^(2x)

Answer 18

another name for "antiderivative" is "indefinite integral"

Answer 19

lets do few more examples
so that you see how to to guess these

Answer 20

whats the derivative of \(\sin x\) ?

Answer 21

cos x

Answer 22

so an antiderivative of \(\cos x\) is \(\sin x\) :

\(\int \cos x\,dx = \sin x+C\)

Answer 23

-sinx +c?

Answer 24

its +sinx


because the derivative of \(\sin x\) is \(\cos x\),
the antiderivative of \(\cos x\) is \(\sin x+C\)

Answer 25

ok got it

Answer 26

lets do couple more

find an antiderivative of \(\large x^2\)

Answer 27

x^3/3

Answer 28

correct. but why ?

Answer 29

the formula is x^n+1/n+1 ?

Answer 30

nope, don't use the formula

use the fact that differentiating x^3/3 gives you x^2,
so x^3/3 is an antiderivative of x^2

Answer 31

ok

Answer 32

find an antiderivative of \(e^{2x}\)

Answer 33

that makes sense ill just from now on relate that

Answer 34

in other words, what function do u need to differentiate in order to produce \(e^{2x}\)

Answer 35

I was considering how: \[
\left(\frac{u}{v}\right)' = \frac{u'v-uv'}{v^2}
\]I put \(v=1+2x\), and then got the equation: \[
(1+2x)u' - (2)u = xe^{2x}
\]I divide both sides by \(x\) and got: \[
\frac{u'-2u}{x} +2u' = e^{2x}
\]I came to the impression that if I can get \(u'=2u\), I could make things simpler.
So I used \(u=Ce^{2x}\).\[
\frac{2Ce^{2x}-4Ce^{2x}}{x} +4Ce^{2x} = e^{2x} \implies 4Ce^{2x} =e^{2x} \implies C=\frac 14
\]That got me to my solution of\[
\frac uv = \frac{\frac{1}{4}e^{2x}}{1+2x} = \frac{e^{2x}}{4+8x}
\]This isn't a reliable method, but I think it's one of the few cases where you can reverse engineer the quotient rule.

Answer 36

i cant do it that way though thats cool

Answer 37

(e^(2x))/2

Answer 38

also wio that is the correct answer

Answer 39

v=e^2/2

Answer 40

Nice! as you can see integration is indeed a guessing game and you can only get good at it by practice

Answer 41

are you saying (e^(2x))/2 is an antiderivative of e^(2x) ?

Answer 42

yes

Answer 43

why

Answer 44

cause the derivative of that is e^2x

Answer 45

Perfect! i wanted to hear that from you :)

Answer 46

so you can work the rest of the problem