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Math Help - Integration by substitution

  1. #1
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    Integration by substitution

    Hello, a little stuck on a question:
    Integrate (x+1)(x+3)^5 u = (x+3)

    I have ended up with 1/7(x+3)^7 - 1/3(x+3)^6 + K

    but the answer is 1/21 (x+3)^6(3x+2) ...

    Can someone show me how to do it please

    Thanks!
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    You integrate the thing and then you will get

    From your answer take (x+3)^6 out as common factor

    You will get

    (x+3)^6 [ 1/7(x+3) - 1/3] + K


    Any more help ?
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  3. #3
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    i must have made a mistake somewhere else because although thats very useful and what i was missing (thank you ) the answer i now get is:

    21(x+3)^6(3x+2)

    whereas the answer is:

    \frac{1}{21} (x+3)^6(3x+2) ... ??
    Last edited by darksupernova; April 20th 2010 at 02:37 AM.
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  4. #4
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    Quote Originally Posted by darksupernova View Post
    i must have made a mistake somewhere else because although thats very useful and what i was missing (thank you ) the answer i now get is:

    21(x+3)^6(3x+2)

    whereas the answer is:

    1/21 (x+3)^6(3x+2) ... ??
     \frac{(x+3)^7}{7} - \frac{(x+3)^6}{3}

     (x+3)^6 \{ \frac{(x+3)}{7} - \frac{1}{3} \}

     (x+3)^6 \{ \frac{3(x+3)}{7 \cdot 3} - \frac{7}{3 \cdot 7 } \}

     (x+3)^6 \{ \frac{3(x+3)-7}{21} \}

     (x+3)^6 \{ \frac{3x+9-7}{21} \}

     (x+3)^6 \{ \frac{(3x+2)}{21} \}

    That's it!

    Feel free to ask if you still find it difficult ..
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  5. #5
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    ahh... didnt think about doing that!! thanks!
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  6. #6
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    Quote Originally Posted by darksupernova View Post
    Hello, a little stuck on a question:
    Integrate (x+1)(x+3)^5 u = (x+3)

    I have ended up with 1/7(x+3)^7 - 1/3(x+3)^6 + K

    but the answer is 1/21 (x+3)^6(3x+2) ...

    Can someone show me how to do it please

    Thanks!
    Hi darksupernova,

    Using your substitution u=x+3, so x+1=u-2, du=dx

    \int{(u-2)u^5}du=\int{\left(u^6-2u^5\right)}du=\frac{u^7}{7}-\frac{u^6}{3}+C

    =\frac{3u^7-7u^6}{21}+C

    =\frac{u^6(3u-7)}{21}+C=\frac{(x+3)^6[3(x+3)-7]}{21}+C

    =\frac{(x+3)^6(3x+9-7)}{21}+C=\frac{(x+3)^6(3x+2)}{21}+C
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