## chrisplusian 3 years ago Quick integration question

1. chrisplusian

$\int\limits x-1+\frac{ 1 }{ x+2 } +\frac{ 1 }{ x-1 } dx$

2. chrisplusian

is this the same as:

3. chrisplusian

$\int\limits x-1 dx +\int\limits \frac{ 1 }{ x+2 } dx +\int\limits \frac{ 1 }{ x-1 } dx$ ????

4. AccessDenied

Yes, that would be correct.

5. chrisplusian

I am doing integration by partial fraction decomposition, I got this far on my own but when I checked wolfram alpha it showed the "x-1" as the integral of x dx, and then the integral of -1 dx and somehow their answer and my own do not match

6. AccessDenied

You may also split (x - 1) into separate integrals, that should work out as well.

7. chrisplusian

8. chrisplusian

$x^2 -2x +1 +2\ln \left| x+2 \right| +2\ln \left| x-1 \right|$

9. chrisplusian

and

10. chrisplusian

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

11. chrisplusian

??

12. chrisplusian

The first one is my answer the second is wolfram alpha

13. AccessDenied

Your answer appears to be about the same as wolfram's, except it was multiplied by 2, which doesn't work. The absolute values are fine to have here. Also, since you are doing an indefinite integration, both answers would have an additional arbitrary constant.

14. chrisplusian

when I had the integral of x-1 dx I substituted and said u=x-1 and du=dx. then integrating I ended up with (u^2)/2. plugging back in for u I got the polynomial over two. to clear the fraction from the entire thing I multiplied each term by 2 hence the 2ln.... is that wrong?

15. AccessDenied

yeah, that wouldn't work out because you essentially changed the value of the original integral. Multiplying by 2 also multiplies your original integral all by 2.

16. chrisplusian

I can see how my answer is similar to wolfram's but, when you distribute the x into the (x-2) of wolframs solution you get (x^2)-2x an that is all divided by two. My solution was (X^2)-2x +1.

17. chrisplusian

so you are saying that I can't take$\frac{ x^2 -2x +1 }{ 2 } +\ln \left| x+2 \right| +\ln \left| x-1 \right| +c$

18. chrisplusian

and multiply through by two to clear the fraction?

19. AccessDenied

Yes, since you basically have: $\int (x - 1 + \frac{1}{x+2} + \frac{1}{x-2}) \; dx = \frac{1}{2}(x^2 - 2x + 1) + \ln|x + 2| + \ln|x-1| + c$ You'd have to multiply both sides by 2 for the equality to hold.

20. AccessDenied

So, you'd have a slightly different integral. However, note that your answers vary only by a constant when we don't multiply by 2 (if we add in absolute values to their answer): 1/2(x^2 - 2x) + ln|x+2| + ln|x-1| + c1 = 1/2(x^2 - 2x + 1) + ln|x+2| + ln|x-1| + c2 1/2(x^2 - 2x) + ln|x+2| + ln|x-1| + c1 = 1/2(x^2 - 2x) + ln|x+2| + ln|x-1| + 1/2 + c2 c1 = 1/2 + c2 <-- They only vary by constant values Since c2 is just some random constant, we could also just combine 1/2 + c2 into one constant.

21. chrisplusian

ok that is kind of confusing. so my answer was right just without the minus two. Is there a way I can use wolfram to see if my answer is equivalent to the one they give me?

22. chrisplusian

I meant times two

23. Chlorophyll

= x²/2 - x + ln | x+2| + ln | x -1| + C

24. Chlorophyll

or x² - 2x + 2ln | x+2| + 2ln | x -1| + C

25. Chlorophyll

You have 4 separates terms to take derivative. The result looks neat without fraction !

26. Chlorophyll

*separated

27. chrisplusian

I am still somewhat confused I don't get how they are the same

28. Chlorophyll

The first one has fraction 1/2 If you don't want fraction form, you need to multiply all terms by 2!

29. Chlorophyll

= x²/2 - x + ln | x+2| + ln | x -1| + C x² - 2x + 2ln | x+2| + 2ln | x -1| + C

30. AccessDenied

How would you multiply by 2 without changing the value of the original integral? If you take the derivative of the indefinite integral multiplied by 2, you get the original function, multiplied by 2...

31. Chlorophyll

This is how you multiply: = 2 ( x²/2 - x + ln | x+2| + ln | x -1| ) + C = x² - 2x + 2ln | x+2| + 2ln | x -1| + C

32. chrisplusian

ok to start with I did that but then @AccessDenied said I could not. The basic problem I have is understanding how wolfram alpha's answer is the same as mine. @Chlorophyll could you scroll up to where i posted my answer as compared to my answer?

33. chrisplusian

I got (x^2)-2x+1 and I just cannot see how that is the same as (x-2)x

34. Chlorophyll

This is just plain straight forward formula from the textbook!

35. Chlorophyll

" I got (x^2)-2x+1 " Where do you get 1?

36. chrisplusian

sorry I must have missed the "plain straightforward" part of my calculus II textbook

37. chrisplusian

lol

38. Chlorophyll

Does it mean you understand your mistake?

39. AccessDenied

It is an indefinite integral, so your answer has one constant of variation added on. The difference between your answers is a constant, so it can be combined with your constant of variation since C + 1/2 is also a constant. :P

40. chrisplusian

I used the "x-1" as u then du=dx. so I then have the integral of u du. That evaluated becomes (u^2)/2. plug x-1 back in for u and then square it. that gives you (x^2)-2x+1 all divided by two

41. Chlorophyll

u = x -1 --> du = dx => ∫ du/ u = ln u = ln | x-1|

42. chrisplusian

it was just x-1 not 1/(x-1)

43. Chlorophyll

Oh, so they're 2 individual terms!

44. chrisplusian

can't you integrate them as I just described? I just ask cause that is where my mind went when I had to do it without any help

45. Chlorophyll

No, it doesn't work that way. We only apply substitution when we're unable to integral term by term: Suppose ∫ ( x - 1) dx Let u = x -1 => du = dx =>∫ udu = u²/2 = ( x-1)² /2 = ( x² - 2x + 1 ) /2 = x²/2 - x + 1/2

46. Chlorophyll

* to integrate

47. AccessDenied

= x²/2 - x + 1/2 [+ c] Let 1/2 + c = C (combine constants into a single constant of variation) = x²/2 - x + C

48. AccessDenied

The big thing about indefinite integration is that there is an additional constant of variation due to differentiation removing any constants and you are reversing the process. These answers only vary by constants, where differentiation removes both 1/2 and your constant of variation...