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Math Help - Solve the following integral:

  1. #1
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    Solve the following integral:

    \int{(x^2 + 5x + 6)\sqrt{x + 1}}dx by the addition-subtraction method.


    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; January 14th 2011 at 01:54 PM.
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  2. #2
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    Quote Originally Posted by wonderboy1953 View Post
    \int{(x^2 + 5x + 6)\sqrt{x + 1}}dx by the addition-subtraction method.
    Perhaps you can elaborate on this method since a google search on this technique provides nothing useful.

    integration addition subtraction method - Google Search
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Perhaps you can elaborate on this method since a google search on this technique provides nothing useful.

    integration addition subtraction method - Google Search
    I'll refer you to this thread "The addition-subtraction puzzle (calculus)" in the puzzle section.
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  4. #4
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    I'm sort of confused here. What are you trying to achieve with this integral?
    It's an absolute no-brainer. I'm just as confused about this alleged 'method'!

    Letting t = \sqrt{x+1} gives:

    \begin{aligned}& \int \left(x^2+5x+6\right)\sqrt{x+1}\;{dx}  = \int {2(t^2+1)(t^2+2)t^2\;{dt} \\& = \int (2 t^6+6 t^4+4 t^2)\;{dt} =  \frac{2}{7}t^7+\frac{6}{5}t^5+\frac{4}{3}t^3+k \\& = \frac{2}{7}\sqrt{(x+1)^7}+\frac{6}{5}\sqrt{(x+1)^5  }+\frac{4}{3}\sqrt{(x+1)^3}+k.\end{aligned}
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  5. #5
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    Quote Originally Posted by TheCoffeeMachine View Post
    I'm sort of confused here. What are you trying to achieve with this integral?
    It's an absolute no-brainer. I'm just as confused about this alleged 'method'.

    Letting t = \sqrt{x+1} gives:

    \begin{aligned}& \int \left(x^2+5x+6\right)\sqrt{x+1}\;{dx}  = \int {2(t^2+1)(t^2+2)t^2\;{dt} \\& = \int (2 t^6+6 t^4+4 t^2)\;{dt} =  \frac{2}{7}t^7+\frac{6}{5}t^5+\frac{4}{3}t^3+k \\& = \frac{2}{7}\sqrt{(x+1)^7}+\frac{6}{5}\sqrt{(x+1)^5  }+\frac{4}{3}\sqrt{(x+1)^3}+k.\end{aligned}
    I didn't even bother to look it up in the forum.
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  6. #6
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    Please recheck your algebra

    Quote Originally Posted by TheCoffeeMachine View Post
    I'm sort of confused here. What are you trying to achieve with this integral?
    It's an absolute no-brainer. I'm just as confused about this alleged 'method'!

    Letting t = \sqrt{x+1} gives:

    \begin{aligned}& \int \left(x^2+5x+6\right)\sqrt{x+1}\;{dx}  = \int {2(t^2+1)(t^2+2)t^2\;{dt} \\& = \int (2 t^6+6 t^4+4 t^2)\;{dt} =  \frac{2}{7}t^7+\frac{6}{5}t^5+\frac{4}{3}t^3+k \\& = \frac{2}{7}\sqrt{(x+1)^7}+\frac{6}{5}\sqrt{(x+1)^5  }+\frac{4}{3}\sqrt{(x+1)^3}+k.\end{aligned}
    A no-brainer for you and a few others perhaps (I could have given a much tougher problem). The A-S method is simpler/more efficient to apply to this type of problem, yet I've never seen it used in any math class nor text.

    A simpler method would help avoid algebraic mistakes (btw please recheck your algebra as it needs fixing).

    Since the A-S method hasn't been applied, my challenge still stands. You can try again TheCoffeeMachine or maybe someone else out there wants to take a whack at this problem (refer to the third post on this thread).

    Since the A-S method hasn't been used to solve this problem, my challenge still stands and if you want to see details about this method, I'll just refer you again to the third post on this thread.
    Last edited by wonderboy1953; January 15th 2011 at 10:25 AM. Reason: incomplete sentence
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    Quote Originally Posted by wonderboy1953 View Post
    A simpler method would help avoid algebraic mistakes (btw please recheck your algebra as it needs fixing).
    Where? I've rechecked and I can't find anything that needs fixing! In fact, it's perfect!
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    Quote Originally Posted by TheCoffeeMachine View Post
    Where? I've rechecked and I can't find anything that needs fixing! In fact, it's perfect!
    Should be \int {(t^2+1)(t^2+2)t\;{dt}, not \int {2(t^2+1)(t^2+2)t^2\;{dt}
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  9. #9
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    Quote Originally Posted by wonderboy1953 View Post
    Should be \int {(t^2+1)(t^2+2)t\;{dt}, not \int {2(t^2+1)(t^2+2)t^2\;{dt}
    If you differentiate his answer you get the integrand.
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    Quote Originally Posted by wonderboy1953 View Post
    Should be \int {(t^2+1)(t^2+2)t\;{dt}, not \int {2(t^2+1)(t^2+2)t^2\;{dt}
    Wrong! Let me show you in slower/detailed steps, my dear wonderboy.

    We have t = \sqrt{x+1} \Rightarrow \frac{dt}{dx} = \frac{1}{2\sqrt{x+1}} \Rightarrow dx = 2\sqrt{x+1}\;{dt} = 2t\;{dt}. Similarly, we have
    t = \sqrt{x+1} \Rightarrow t^2 = x+1 and x^2+5x+6 = (x+2)(x+3) = (t^2+1)(t^2+2).

    Thus \int \left(x^2+5x+6\right)\sqrt{x+1}\;{dx} = \int {2(t^2+1)(t^2+2)t^2\;{dt}.
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    Step-by-step explanation

    Quote Originally Posted by chiph588@ View Post
    If you differentiate his answer you get the integrand.
    TheCoffeeMachine substituted t = \sqrt{x + 1}

    This means that x = t^2 - 1

    When you do the substitution in the starting equation, you have:

    \int{[(t^2 - 1)^2 + 5(t^2 - 1) + 6]t\;{dt}} which leads to

    \int {(t^2+1)(t^2+2)t\;{dt}
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  12. #12
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    Oops

    I overlooked the differential so TheCoffeeMachine is right.
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  13. #13
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    Hint

    Quote Originally Posted by wonderboy1953 View Post
    \int{(x^2 + 5x + 6)\sqrt{x + 1}}dx by the addition-subtraction method.


    Moderator edit: Approved Challenge question.
    For those who may want to see what I have in mind, you can factor the (x^2 + 5x + 6) portion into (x + 2)(x + 3). With the first factor, you can subtract and add 1 to change it into [(x + 1) + 1] (do you see why?) What would you do with the second factor to go on to complete the problem using the addition-subtraction method?
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