Math question

I usually don’t need to ask these any more, but I’m not seeing how one step proceeds to the next.

xln(2x+1) – x + 1/2ln(2x+1) + C

becomes, in the solutions manual:

1/2(2x+1)ln(2x+1) – x + C

Where did the x go? How is it that ln is only in front of one of the (2X+1) terms? How did it go from adding to multiplying? ln rule of lnx+lny = lnxlny, but why is there only an ln in front of one of the terms? What hte heck happened there? Thanks.

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A mind weapon riding along with Four Horsemen of the Apocalypse.https://en.gravatar.com/profiles/edit/#
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13 Responses to Math question

  1. Ryu says:

    Well,
    xln(2x+1) = ln (2x+1)^x
    and
    .5ln(2x+1) = ln (2x+1)^.5
    so add those terms.
    and
    1/2 (2x+1) = x + 1/2
    Looks like they just combined terms and moved the exponent back in front.

    • mindweapon says:

      Still not seeing it. How do you add:

      xln(2x+1) = ln (2x+1)^x
      and
      .5ln(2x+1) = ln (2x+1)^.5?

      Wouldn’t you get ln(2x+1)^.5+x?

      Where did the x go? Why is ln only on one of the terms and not the other?

  2. Steve Johnson says:

    Easy once you look at it as an equation.

    x ln(2x+1) – x + 1/2 ln(2x+1) + C = 1/2 (2x+1) ln(2x+1) – x + C

    + x – C to both sides

    x ln(2x+1) + 1/2ln(2x+1) = 1/2(2x+1)ln(2x+1)

    Factor out a ln(2x+1)

    (x + 1/2) ln (2x+1) = 1/2(2x+1) ln(2x+1)

    Multiply the left side by 2/2

    1/2*2(x+1/2) ln (2x+1) = 1/2(2x+1) ln(2x+1)

    1/2(2x+1) ln (2x+1) = 1/2(2x+1) ln(2x+1)

  3. Piotr says:

    xln(2x+1) – x + 1/2ln(2x+1) + C = [ln(2x + 1)] (x + 1/2) – x + C = 1/2(2x+1)ln(2x+1) – x + C

  4. FD says:

    Why doesn’t the book leave it as (x+1/2)ln(2x+1) – x + C? Is the next step integration by parts?

  5. countenance says:

    Look at all ths right wing ignorance on parade.

    I bet this equation is for making some sort of secret potion that takes away the ability for black people to build pyramids and be kings and queens.

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