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Math Help - Integration help required~~

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    Integration help required~~

    Integration help required~~-2.jpgIntegration help required~~-4.jpgIntegration help required~~-5.jpg

    Could someone please help me integrate these functions? :S
    I've tried them for hours and hours but my answers are always away by some numbers ...

    Thank you in advance!!

    ps. they are three different individual functions..
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    Re: Integration help required~~

    Quote Originally Posted by ghfjdksla94 View Post
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    Could someone please help me integrate these functions? :S
    I've tried them for hours and hours but my answers are always away by some numbers ...

    Thank you in advance!!

    ps. they are three different individual functions..
    First one:

    \displaystyle \begin{align*} \int{ \left( \frac{4.8}{1 + 6e^{-0.46x}} \right)^2 \,dx} &= \int{ \frac{ -\frac{50}{23} e^{ -\frac{23}{50} x } }{ -\frac{50}{23} e^{-\frac{23}{50} x} } \left( \frac{4.8}{1 + 6e^{ -\frac{23}{50} x} } \right)^2 \,dx} \end{align*}

    Now make the substitution \displaystyle \begin{align*} u = e^{-\frac{23}{50}x} \implies du = -\frac{50}{23}e^{-\frac{23}{50}x}\,dx \end{align*} and the integrate using Partial Fractions.
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    Re: Integration help required~~

    Thanks! But sorry to bother you .. but could you show me some more working outs...? Ive been staring at the help you gave me and trying for half an hour but im getting nowhere....
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    Re: Integration help required~~

    Quote Originally Posted by ghfjdksla94 View Post
    Thanks! But sorry to bother you .. but could you show me some more working outs...? Ive been staring at the help you gave me and trying for half an hour but im getting nowhere....
    When you make that substitution you should get \displaystyle \begin{align*} 10.5984\int{ \frac{1}{u(1 + 6u)^2}\,du } \end{align*}. Now apply Partial Fractions
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    Re: Integration help required~~

    For the second problem, expand the integrand, then integrate term by term.

    The exponential terms may be integrated by beginning with the derivative:

    \frac{d}{dx}\left(4^{kx} \right)=k\ln(4)4^{kx} hence:

    \frac{1}{k\ln(4)}\int\,d\left(4^{kx} \right)=\int 4^{kx}\,dx

    \int 4^{kx}\,dx=\frac{4^{kx}}{k\ln(4)}+C
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    Re: Integration help required~~

    Thank you again but now im sort of getting embarassed because I never learnt any partial fractions, so even though I looked at series of websites, that explains partial fraction, I still can't apply it... I know it's really annoying, but could you help me just a little bit more???
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    Re: Integration help required~~

    Quote Originally Posted by ghfjdksla94 View Post
    Thank you again but now im sort of getting embarassed because I never learnt any partial fractions, so even though I looked at series of websites, that explains partial fraction, I still can't apply it... I know it's really annoying, but could you help me just a little bit more???
    Try \displaystyle \begin{align*} \frac{A}{u} + \frac{B}{1 + 6u} + \frac{C}{(1 + 6u)^2} \equiv \frac{1}{u(1 + 6u)^2} \end{align*}

    Try to simplify the LHS and evaluate A, B, C.
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    Re: Integration help required~~

    You begin by assuming the integrand may be expressed in the form:

    \frac{1}{u(6u+1)^2}=\frac{A}{u}+\frac{B}{6u+1}+ \frac{C}{(6u+1)^2}

    Now multiply through by the lowest common denominator u(6u+1)^2

    1=A(6u+1)^2+Bu(6u+1)+Cu

    Now, expand the right side and arrange as a quadratic in standard form, and write the left side as:

    0u^2+0u+1

    Then equate corresponding coefficients to arrive at a linear 3X3 system in A, B, and C for which there is a unique solution.
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    Re: Integration help required~~

    Thank you for your help once again, but would it be asking too much if I ask you to solve this problem (the second one) for me? I know you are only meant to give directions and helps but I have a very limited knowledge about partial fractions. Also, I have like 5 of questions that look similar but with completely different numbers, and other 10-15 questions completely different. So could you please do me a favour just once?
    Last edited by ghfjdksla94; September 16th 2012 at 02:11 AM.
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    Re: Integration help required~~

    Quote Originally Posted by ghfjdksla94 View Post
    Thank you for your help once again, but would it be asking too much if I ask you to solve this problem (the second one) for me? I know you are only meant to give directions and helps but I have a very limited knowledge about partial fractions. Also, I have like 5 of questions that look similar but with completely different numbers, and other 10-15 questions completely different. So could you please do me a favour just once?
    Absolutely not, it is YOUR work to do, not ours. Show what you have tried and where you are stuck, to prove you have shown some effort, then we will give you more guidance.
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    Re: Integration help required~~

    Totally understood. What I have done so far is from your advice and MarkFL2's advice. what I have got is
    1=36Au^2+12Au+A+6Bu^2+Bu+Cu

    which is basically an expansion of 1=A(6u+1)^2+Bu(6u+1)+Cu.

    I have no idea how to go from here..
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    Re: Integration help required~~

    You want to arrange both sides of the equation in standard form as quadratics, i.e,:

    0u^2+0u+1=(36A+6B)u^2+(12A+B+C)u+A

    Now, equating the corresponding coefficients on both sides, will give you 3 equations, from which you may determine A, B and C.
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    Re: Integration help required~~

    I started off assuming A=1 as they are corresponding and what I got is A=1, B=-6 and C= 6... Is this right? And if it is, how do I use this??
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    Re: Integration help required~~

    You have two of the values correct. Did you try substituting the solutions into the equations to make sure all 3 equations are satisfied?

    Once you have the correct solutions, then you may put their values into:

    \frac{1}{u(6u+1)^2}=\frac{A}{u}+\frac{B}{6u+1}+ \frac{C}{(6u+1)^2}

    to get the partial fraction decomposition of the original integral. You will then have 3 terms which you can easily integrate.
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    Re: Integration help required~~

    Quote Originally Posted by ghfjdksla94 View Post
    I started off assuming A=1 as they are corresponding and what I got is A=1, B=-6 and C= 6... Is this right? And if it is, how do I use this??
    That is correct, so that tells you \displaystyle \begin{align*} \frac{1}{u(1 + 6u)^2} \equiv \frac{1}{u} - \frac{6}{1 + 6u} + \frac{6}{(1 + 6u)^2} \end{align*}, so now you can integrate this term by term...
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