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Math Help - Integration

  1. #1
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    Integration

    im not sure how to do these, i tried integration by parts but just got stuck or got an unsure answer.. (S= integral)

    1. S p^5 ln p dp

    i got ln p(1/6p^6) - 37/100 p^12 + C, but i think that is wrong

    2. S 1 (x^2 + 1) e^-x dx
    0

    i got 1+ 3e^-1 + C

    3. S b =sqrt(3) arctan (1/x) dx
    a = 1

    4. S b = (1) (r^3)/sqrt (4+r^2)) dr
    a = 0
    Last edited by skabani; September 15th 2008 at 02:29 PM.
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  2. #2
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    Quote Originally Posted by skabani View Post
    im not sure how to do these, i tried integration by parts but just got stuck or got an unsure answer.. (S= integral)

    1. S p^5 ln p dp
    \int x^5 \ln x dx = \int e^{6\ln x} \ln x (\ln x)' dx
    Let t=\ln x.
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  3. #3
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    I think the answer for question 2 should be -1 - 2e^{-1}.
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  4. #4
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    did u go through integration through seperation twice? or did you use a different method/
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  5. #5
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    Quote Originally Posted by skabani View Post
    did u go through integration through seperation twice? or did you use a different method/
    First I broke the integral apart: \int _0 ^1 x^2e^{-x} + e^{-x} dx = \int _0 ^1 x^2e^{-x} dx + \int _0 ^1 e^{-x} dx

    Then I used integration by parts twice on the first integral to yield -x^2e^{-x} - 2xe^{-x} -2 \int _0 ^1 e^{-x} dx.

    Adding the second integral gives -x^2e^{-x} - 2xe^{-x} - \int _0 ^1 e^{-x} dx = -x^2e^{-x} - 2xe^{-x} + e^{-x} (evaulated from 0 to 1).
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  6. #6
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    thanks.

    do you happen to know what to do for number 4, because i am completely stuck there...
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  7. #7
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    Try using a trig substitution. I solved the problem using r = 2 \tan \theta.
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by skabani View Post
    im not sure how to do these, i tried integration by parts but just got stuck or got an unsure answer.. (S= integral)

    ...
    [snip]

    4. S b = (1) (r^3)/sqrt (4+r^2)) dr
    a = 0
    Make the trig substitution r=2\tan\vartheta\implies \,dr=2\sec^2\vartheta\,d\vartheta

    Thus, the integral becomes \int_0^{\tan^{-1}\left(\frac{1}{2}\right)}\frac{16\tan^3\vartheta  \cdot\sec^2\vartheta\,d\vartheta}{4\sec\vartheta}=  4\int_0^{\tan^{-1}\left(\frac{1}{2}\right)}\tan^3\vartheta\sec\var  theta\,d\vartheta

    (Hint: to integrate this, break off a factor of \tan\vartheta and then apply the identity \tan^2\vartheta=\sec^2\vartheta-1; then apply the substitution z=\sec\vartheta)

    Try to take it from here

    --Chris
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    \int x^5 \ln x dx = \int e^{6\ln x} \ln x (\ln x)' dx
    Let t=\ln x.

    im not sure how to do this still, i keep getting different anwers every time I try it by integration by parts..can you help me out?
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