anonymous one year ago By expressing the following in partial fractions evaluate the given intergral.

1. anonymous

|dw:1437741496765:dw|

2. UsukiDoll

so first we have to take partial fractions before we can take the antiderivative. we can ignore the integral symbol for the first half of this problem

3. UsukiDoll

now there are two forms of partial fractions: factored and non-factored we have to look closely in the denominator.. if we can factor something out. our numerator will just be a single letter like A for example suppose we are given this fraction $\frac{1}{x^2-1}$ we can factor $x^2-1 = (x+1)(x-1)$ $\frac{1}{(x+1)(x-1)}$ so for each partial fraction we have $\frac{A}{x+1} +\frac{B}{x-1}$

4. UsukiDoll

now we have to take the LCD which is (x-1)(x+1) $\frac{x-1}{x-1} \cdot \frac{A}{x+1} +\frac{B}{x-1} \cdot ]\frac{x+1}{x+1}$

5. anonymous

I dont understand how u got from 1/(x^3+x) to 1/(x^2-1) the rest u wrote below, Im familiar with.

6. UsukiDoll

I am providing an example first anyway then we distribute the A and B $\frac{Ax-A+Bx+B}{(x-1)(x+1)}$

7. UsukiDoll

so we have these equations -A+B = 1 A+B = 0 we got the one from the numerator of the original equation

8. anonymous

ah ok that makes sense.

9. UsukiDoll

so now we solve these system of equations through elimination method. That would give us 2B=1 or B =1/2 so we plug that value in one of the equations $A+\frac{1}{2}=0$ so our $A = \frac{-1}{2}$

10. UsukiDoll

yes I like to give a guideline first before we go to the actual problem

11. anonymous

that example makes actual sense, my book skipped alot of steps u took.

12. UsukiDoll

oh dear T_T glad I came lol so we plug in A = -1/2 and B = 1/2 back into the equation before we took the LCD $\large \frac{\frac{-1}{2}}{x+1} +\frac{\frac{1}{2}}{x-1}$

13. UsukiDoll

then we take the antiderivative which is just logs

14. UsukiDoll

$\frac{1}{2} \ln(x+1)-\frac{1}{2}\ln(x-1) +C$

15. anonymous

x+1 is under -1/2 dont u mean vice versa with the pos/negative signs on the 1/2s?

16. UsukiDoll

oh I missed the sign sorry

17. UsukiDoll

$\frac{-1}{2} \ln(x+1)-\frac{1}{2}\ln(x-1) +C$

18. UsukiDoll

so let's apply what we've done to $\large \frac{1}{x^3+x}$

19. UsukiDoll

can we factor the denominator?

20. anonymous

x(x^2+1)

21. UsukiDoll

woo hoo! that means we can use single letters ^_^ should it be non-factorable we have to use Ax+B :(

22. UsukiDoll

if I make typos, please stop me. It's almost 3 am and I am typing fast

23. UsukiDoll

so anyway

24. anonymous

Haha, okey :)

25. UsukiDoll

$\large \frac{1}{x(x^2+1)}$

26. UsukiDoll

so since it's factorable we use single letters $\frac{A}{x}+\frac{B}{x^2+1}$

27. UsukiDoll

so now we use LCD

28. UsukiDoll

$\frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{B}{x^2+1} \cdot \frac{x}{x}$

29. anonymous

cant I just skip that and write A(x^2+1)+bx

30. UsukiDoll

then we distribute $\frac{Ax^2+A+Bx}{x(x^2+1)}$ I don't know if it's a good idea though. skipping steps at this topic is not recommended in my opinion

31. anonymous

okey, I will follow through then:)

32. UsukiDoll

hmmm ... the thing is we have these equations A = 1 Bx = 0 Ax^2 = 0

33. IrishBoy123

you have to use: $$\large \frac{A}{x}+\frac{Bx + C}{x^2+1}$$

34. IrishBoy123

you have a quadratic on the bottom of the second term so you need a linear term on top

35. UsukiDoll

ah geez... no problem... OH that's right x^2+1 doesn't factor thanks for the head up

36. UsukiDoll

yeah x^2+1 doesn't factor so Bx+C so $x^3+x \rightarrow (x)(x^2+1)$ a portion is factored. we can take an x out but we can't do anything to x^2+1

37. UsukiDoll

$\large \frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{Bx + C}{x^2+1} \cdot \frac{x}{x}$

38. UsukiDoll

there we go now we can distribute $\large \frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}$

39. UsukiDoll

now ...... we just have one value from the original equation $\large \frac{1}{x^3+x}$ So we have three equations A = 1 C =0 A+B = 0

40. UsukiDoll

you're probably wondering why 1? why is the rest of the equations = 0 it's because our numerator is just that number if we have something like 1 + 2x on the numerator the C = 2 due to the fact that we had Cx after we did partial fractions

41. UsukiDoll

It's like this A= 1 Cx = 0 Ax^2+Bx^2 = 0 because there is no number next to the x and no number next to the x^2 so they are set to 0... we don't have to write the variable of x's for these letter equations

42. anonymous

ohh okey

43. UsukiDoll

so we know that A = 1 C = 0 A+B = 0 we know C and A so we solve for B 1+B=0 B = -1

44. UsukiDoll

so we plug in A = 1, B =-1, and C = 0 back into the corrected equation shout out to @IrishBoy123 !!!!!!!! WOOOOOOOOOOOOOOOO

45. UsukiDoll

$\large \frac{A}{x}+\frac{Bx + C}{x^2+1}$ $\large \frac{-1}{x}+\frac{1x + 0}{x^2+1}$ $\large \large \frac{-1}{x}+\frac{x}{x^2+1}$

46. UsukiDoll

now we can take antiderivatives... finally after that long journey we CAN INTEGRATE!

47. UsukiDoll

first one is just ln x $\large - \ln x+\frac{x}{x^2+1}$

48. UsukiDoll

the second one is u-substitution

49. IrishBoy123

A = 1, B =-1

50. UsukiDoll

yeah I got that @IrishBoy123

51. UsukiDoll

A =1 , B =-1, and C = 0

52. UsukiDoll

but thanks for the verification :)

53. anonymous

$\frac{ -1 }{ x^2 +1} is...?$

54. UsukiDoll

so back to u-substitution $\large u=x^2+1$

55. UsukiDoll

$\large - \ln x+\frac{x}{x^2+1}$ u-subsitution $\large u=x^2+1$ $\large du = 2x dx$ $\large \frac{1}{2}du = x dx$ $\large - \ln x+(\frac{1}{2})\frac{1}{u}$

56. UsukiDoll

the antiderivative of 1/u is ln u

57. UsukiDoll

$\large - \ln x+(\frac{1}{2}) (\ln u) +C$ then subsitute u back in which was $\large u=x^2+1$ $\large - \ln x+(\frac{1}{2}) (\ln x^2+1) +C$

58. UsukiDoll

whew

59. UsukiDoll

oops miss the l l $\large - \ln \mid x \mid +(\frac{1}{2}) (\ln \mid x^2+1 \mid) +C$

60. UsukiDoll

so to check we just take the derivative.. it should come out to be what we had before

61. anonymous

U spent over 30min for this question DAAMN. Thanks bro, Your a lifesaver.

62. UsukiDoll

$\large -\frac{1}{x} + \frac{1}{2}\frac{1}{x^2+1}(2x) + \ln \mid x^2+1 \mid(0)$ then we got $\large -\frac{1}{x}+\frac{x}{x^2+1}$

63. UsukiDoll

which is what we had earlier... thanks XD

64. anonymous

yup, got the correct answer as the book asked for. R u a math teacher? or tuitor of any kind?

65. UsukiDoll

math major going into teaching.

66. UsukiDoll

Partial Fractions shows up again in Ordinary Differential Equations.

67. anonymous

I will think of you when they do :D

68. anonymous

Alright closing this, THANK YOU <3