## anonymous one year ago anyone good with integration by parts? Integrate ((x)e^(2x))/(1+2x)^2

1. anonymous

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

2. anonymous

ok ive gotten to where you are

3. anonymous

i mean understanding what you did

4. ganeshie8

small typo, let me fix it quick

5. ganeshie8

Let : $$u=xe^{2x} \implies du = (1\cdot e^{2x}+x\cdot 2e^{2x})dx = e^{2x}(1+2x)dx$$ and $$dv = \dfrac{1}{(1+2x)^2} dx\implies v = -\dfrac{1}{2(1+2x)}$$

6. ganeshie8

\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}

7. ganeshie8

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

8. anonymous

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

9. ganeshie8

are you saying you dont know how to integrate $$e^{2x}$$ ?

10. anonymous

yea

11. anonymous

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

12. ganeshie8

ohkie, its easy, whats the derivative of $$e^{2x}$$ ?

13. anonymous

e^(2x)?

14. ganeshie8

heard of chain rule before ?

15. anonymous

oh

16. anonymous

e^(2x)2

17. ganeshie8

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

18. ganeshie8

another name for "antiderivative" is "indefinite integral"

19. ganeshie8

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

20. ganeshie8

whats the derivative of $$\sin x$$ ?

21. anonymous

cos x

22. ganeshie8

so an antiderivative of $$\cos x$$ is $$\sin x$$ : $$\int \cos x\,dx = \sin x+C$$

23. anonymous

-sinx +c?

24. ganeshie8

its +sinx because the derivative of $$\sin x$$ is $$\cos x$$, the antiderivative of $$\cos x$$ is $$\sin x+C$$

25. anonymous

ok got it

26. ganeshie8

lets do couple more find an antiderivative of $$\large x^2$$

27. anonymous

x^3/3

28. ganeshie8

correct. but why ?

29. anonymous

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

30. ganeshie8

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

31. anonymous

ok

32. ganeshie8

find an antiderivative of $$e^{2x}$$

33. anonymous

that makes sense ill just from now on relate that

34. ganeshie8

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

35. anonymous

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.

36. anonymous

i cant do it that way though thats cool

37. anonymous

(e^(2x))/2

38. anonymous

also wio that is the correct answer

39. anonymous

v=e^2/2

40. ganeshie8

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

41. ganeshie8

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

42. anonymous

yes

43. ganeshie8

why

44. anonymous

cause the derivative of that is e^2x

45. ganeshie8

Perfect! i wanted to hear that from you :)

46. ganeshie8

so you can work the rest of the problem

47. ganeshie8

|dw:1440738169733:dw|

48. anonymous

i still dont know where to go from where we are though. does it go $\frac{ -xe ^{2x}}{ 2( 1+2x)}+uv-\int\limits_{}^{}vdu$

49. ganeshie8

Easy, you just need to plugin $$\int e^{ex}\, dx$$

50. ganeshie8

Easy, you just need to plugin $$\int e^{2x}\, dx$$

51. ganeshie8

maybe let me just show u

52. ganeshie8

\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 \\~\\ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \frac{1}{2} \int e^{2x}\, dx \\~\\ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \frac{1}{2} \dfrac{e^{2x}}{2} + C \\~\\ \end{align}

53. ganeshie8

we're done! simplify if you want to

54. anonymous

the answer is supposed to by e^2x/4(1+2x) + C

55. ganeshie8

you will get that if u simplify

56. ganeshie8
57. anonymous

i simplified and got the answer i needed. THANK YOU SOOO much! i will look that tonight after i study this problem a little more. I am clearly lacking some basic understanding

58. anonymous

I have been working on that problem all literally since like noon to 10 :25 pm you are my new hero

59. anonymous

of course i had completely forgotten about the chain rule for e^2x

60. anonymous

the only thing im really still confused over is how you get -1/2(1+2x) for the first v value?

61. ganeshie8

differentiate -1/2(1+2x) what do you get ?

62. ganeshie8

integration takes some time to get used to... because it requirs some guessing and not so straightforward as differentiation

63. anonymous

does moving the denominator up make it easier? like (1+2x)^-2

64. ganeshie8

|dw:1440739783369:dw|

65. anonymous

oh got it

66. anonymous

i did integral of (1+2x)^-2 = (1+2x)-1/-1

67. anonymous

you have been a huge huge help

68. anonymous

thank you so much for spending so much time with me

69. ganeshie8

looks good! make sure you watch that video from khan academy, it has a very good introduction to integration by parts...

70. anonymous

ok im gonna watch it now. thank you again for you help and your time

71. anonymous

$\int\limits_{?}^{?}\frac{ xe ^{2x}}{ (1+2x)^{2} }; Let U=xe ^{2x}; du=e ^{2x}(1+2x); dv= \frac{ 1 }{ (1+2x)^{2}}; v= -\frac{ 1 }{ 1+2x }$

72. anonymous

$uv-\int\limits_{?}^{?}vdu = -\frac{ xe ^{2x} }{ 2(1+2x) }-\int\limits_{}^{?}-\frac{ 1 }{ 2(1+2x) }*e ^{2x}(1+2x)*dx$

73. anonymous

$= -\frac{ xe ^{2x} }{ 2(1+2x) }-\int\limits -\frac{ 1 }{ 2 }*e ^{2x}dx = -\frac{ xe ^{2x} }{ 2(1+2x)} + \frac{ 1 }{ 2 }\int\limits e ^{2x}dx$

74. anonymous

$-\frac{ xe ^{2x} }{ 2(1+2x) } + \frac{ 1 }{ 2 }*\frac{ e ^{2x} }{ 2 }= -\frac{ xe ^{2x} }{ 2(1+2x)} + \frac{ e ^{2x} }{ 4 } =-\frac{ 4ex ^{2x} }{ (8+16x) }+\frac{ 2e ^{2x} +4ex ^{2x}}{ (8+16x) }$

75. anonymous

$=\frac{ 2e ^{2x} }{ 8(1+2x) } = \frac{ e ^{2x} }{ 4(1+2x) }$