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Math Help - General Calculus.

  1. #1
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    Post General Calculus.

    Hi all!

    I need help on two calculus problems:

    Firstly ...

    y = tan^-^1(lnx)

    I know the solution which is  \frac{dy}{dx} = \frac{1}{x} . \frac{1}{1 + (lnx)^2}

    What are the rules to get this problem solved?

    Secondly ...

    When doing Intergration on partial fraction lets take an example:

    <br />
\int\frac{4x^2 - 24x + 11}{(x-3)^2(x+2)} = \frac{3}{(x+2)} + \frac{1}{(x-3)} - \frac{5}{(x-3)^2}

    Then the integral becomes

    <br />
\int3ln(x+2) + ln(x-3) ... ?

    How can I integrate the last bit:  -\frac{5}{(x-3)^2}

    I hope this will help others too!
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  2. #2
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    Quote Originally Posted by dadon
    Hi all!

    I need help on two calculus problems:

    Firstly ...

    y = tan^-^1(lnx)

    I know the solution which is  \frac{dy}{dx} = \frac{1}{x} . \frac{1}{1 + (lnx)^2}
    This is chain rule.
    You got,
    y=\tan^{-1}(u) where u=\ln x
    Thus,
    \frac{dy}{du}=\frac{1}{1+u^2} and
    \frac{du}{dx}=\frac{1}{x}
    Thus,
    \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}
    Thus,
    \frac{dy}{dx}=\frac{1}{x}\cdot \frac{1}{1+u^2}
    But, u=\ln x thus,
    \frac{1}{x}\cdot \frac{1}{1+\ln^2 x}
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  3. #3
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    Quote Originally Posted by dadon

    How can I integrate the last bit:  -\frac{5}{(x-3)^2}
    You need,
    \int \frac{5}{(x-3)^2}dx
    Let, u=x-3
    Then, du=dx
    Thus, use this substitution to get,
    \int 5u^{-2}du=-5u^{-1}+C
    But, u=x-3 thus,
    -5(x-3)^{-1}+C
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  4. #4
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    Thumbs up re:

    Thank you! I appreciate your help and nicely explained!

    Regards,

    dadon
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