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Math Help - i need help in integration question?

  1. #1
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    i need help in integration question?

    solve this question.......
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by khhratmath123 View Post
    solve this question.......
    Show us your work so far and then we can guide you through where you're stuck.

    As a pointer, the way I would approach this is to factorise the denominator, and then use partial fractions to get two fractions which are easy to integrate.
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by khhratmath123 View Post
    solve this question.......
    Also, if you are having problems factorising the fraction, remember you can take a factor of -1 out, so your problem becomes:

    (-1)\int^{-2}_{-3}\frac{1}{x^2 + 6x+5}dx
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  4. #4
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    i have solved this question but i have doubt....plz check..is it right?
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  5. #5
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    this was easy:

    ∫((dx)/(x+6x+5))
    =∫((dx)/((x+1)(x+5)))

    now use partial fractions ,get the constants and put the limits.

    your answer is definitely wrong as it should come in terms of logs.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by khhratmath123 View Post
    i have solved this question but i have doubt....plz check..is it right?
    Correct.
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  7. #7
    Super Member craig's Avatar
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    Quote Originally Posted by khhratmath123 View Post
    i have solved this question but i have doubt....plz check..is it right?
    No. I see what you have done, but your answer should be:

    [\frac{1}{2}tanh^{-1}(\frac{x+3}{2})]^{-2}_{-3}

    Or the way that me and Pulock suggested, this way you would get the actual logs instead of the inverse hyperbolic.
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by craig View Post
    No.
    \int\frac{dx}{(2)^2-(x+3)^2}

    Let u=x+3\Rightarrow{du}=dx

    \Rightarrow\int\frac{dx}{(2)^2-(x+3)^2}=\int\frac{du}{(2)^2-(u)^2}
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  9. #9
    Super Member craig's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    \int\frac{dx}{(2)^2-(x+3)^2}

    Let u=x+3\Rightarrow{du}=dx

    \Rightarrow\int\frac{dx}{(2)^2-(x+3)^2}=\int\frac{du}{(2)^2-(u)^2}
    I see how you derived that. However, if it was:

    \int\frac{du}{\sqrt{(2)^2-(u)^2}} that would make his answer right.

    But as far as I remember, \int\frac{du}{1-u^2} = tanh^{-1} u ?
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by craig View Post
    I see how you derived that. However, if it was:

    \int\frac{du}{\sqrt{(2)^2-(u)^2}} that would make his answer right.

    But as far as I remember, \int\frac{du}{1-u^2} = tanh^{-1} u ?
    My mind saw a radical where there was none.
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  11. #11
    Super Member craig's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    My mind saw a radical where there was none.
    Easy to do
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  12. #12
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    thanks for all
    i have solved by partial fraction method
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