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Thread: General Calculus.

  1. #1
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    Post General Calculus.

    Hi all!

    I need help on two calculus problems:

    Firstly ...

    $\displaystyle y = tan^-^1(lnx)$

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

    What are the rules to get this problem solved?

    Secondly ...

    When doing Intergration on partial fraction lets take an example:

    $\displaystyle
    \int\frac{4x^2 - 24x + 11}{(x-3)^2(x+2)} = \frac{3}{(x+2)} + \frac{1}{(x-3)} - \frac{5}{(x-3)^2}$

    Then the integral becomes

    $\displaystyle
    \int3ln(x+2) + ln(x-3) ... ?$

    How can I integrate the last bit: $\displaystyle -\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 ...

    $\displaystyle y = tan^-^1(lnx)$

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

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

    Thank you! I appreciate your help and nicely explained!

    Regards,

    dadon
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