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I Integration over something with addition in denominator

  1. Apr 6, 2017 #1
    Hi,
    I'm struggling to figure out how to do integration with forms such as:

    ∫ x/(x+1) dx
    ∫ x/(x+1)^2 dx
    ∫(b-x)^2/(b-a) dx

    This last one especially is giving me a strange issue, where if I plug it into wolfram:
    https://www.wolframalpha.com/input/?i=integrate+(b+-+x)^2+/(b-a)++dx

    It shows up with a result of (b-x)^3/3(b-a) + C
    While if we know a to be a constant 0, and is left out, and this integral is plugged in: ∫(b-x)^2/(b) dx
    https://www.wolframalpha.com/input/?i=integrate+(b+-+x)^2+/(b)++dx

    The answer x^3/3b +bx -x^2 +C is produced

    You get a different result than the result of the former with a =0. The former one ends up with a b^2 term in the end (if you expand the (b-x)^3 and divide by b), while the latter doesn't. What is happening here and how are these types of integrals solved?
     
  2. jcsd
  3. Apr 6, 2017 #2

    jedishrfu

    Staff: Mentor

  4. Apr 6, 2017 #3
    Ah yes, that helped me do the first and second ones here, thanks! On the third I'm still completely stumped though
     
  5. Apr 6, 2017 #4
    ##\displaystyle {1\over b - a}\int (b - x)^2 dx##

    Now substitute ##u = b -x## or just use ##(a+b)^2 = a^2 + b^2 + 2ab##.
     
  6. Apr 6, 2017 #5
    I assume you mean: (b−x)^2 = b^2 + x^2 - 2bx ? That's the route I tried. After integrating that part and multiplying by the 1/(b-a) again, I get:

    (xb^2 + x^3/3 - bx^2)/(b-a) + C

    There's no b^3 term in the numerator there, unlike the answer wolfram provided me. So it seems I'm doing something wrong (assuming wolfram isn't bugged)
     
  7. Apr 6, 2017 #6

    Mark44

    Staff: Mentor

    ##\frac 1{b - a}\int (b - x)^2dx##

    Even simpler than using a substitution, just expand the ##(b - x)^2## expression.
    Your answer above is fine. Possibly you entered it into wolframalpha incorrectly. Could you provide a link to the WA page with your integral?
     
  8. Apr 6, 2017 #7
    https://www.wolframalpha.com/input/?i=integrate+(b+-+x)^2+/(b-a)++dx

    My input in case that link fails:
    integrate (b - x)^2 /(b-a) dx
     
  9. Apr 6, 2017 #8
    You have ##(xb^2+x^3/3-bx^2)/(b-a)+C=\frac{1}{3(b-a)}(3xb^2+x^3-3bx^2)+C##.
    Compare this with the cubic formula ##(x-b)^3=x^3+3xb^2-3bx^2-b^3##
     
  10. Apr 6, 2017 #9
    They are similar, but the latter has a -b^3 in there, which is not a constant. Where is this -b^3 coming from?
     
  11. Apr 6, 2017 #10
    Well, ##b^3## is a constant in the sense that it is independent of ##x##. You can therefore write ##\frac{1}{3(b-a)}(3xb^2+x^3-3bx^2)+C=\frac{1}{3(b-a)}(x-b)^3+\frac{b^3}{3(b-a)}+C=\frac{1}{3(b-a)}(x-b)^3+D##, where ##D=\frac{b^3}{3(b-a)}+C##
     
  12. Apr 6, 2017 #11
    I suppose so, and I guess the term would disappear if you used a definite integral. That sounds like it should be the intended reading of the result here. Thanks!
     
  13. Apr 6, 2017 #12
    If you had used substitution you would have got the Wolfram answer.

    ##\displaystyle {1\over b - a}\int (b - x)^2 dx = {1\over a - b}\int (b-x)^2 d(b - x) = {(b-x)^3/3\over a - b}##.
     
  14. Apr 6, 2017 #13
    Yes, for an indefinite integral the constant is arbitrary. For an definite integral on the interval ##[c,d]## you would have ##F(d)-F(c)##, where ##F(x)## denotes the indefinite integral. So, the constant disappears since you subtract.
     
  15. Apr 7, 2017 #14
    ##F(x)## denotes a antiderivative and indefinite integral is "family" of all antiderivatives. Clearly ##F(x)## does not denote a indefinite integral.

    Correct me if I am wrong.
     
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