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  1. #1
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    integration question

    integrate 1 / (3-2x)^2. i dont understand how to integrate in this format, can someone please help me?
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  2. #2
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    o__O

    Take t=3-2x
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  3. #3
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    Hi,

    Or make appear the derivate of (3-2x) above, and you'll get u'(x)/uČ(x), and its integrate is -1/u
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  4. #4
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    Quote Originally Posted by -sonal- View Post
    integrate 1 / (3-2x)^2. i dont understand how to integrate in this format, can someone please help me?
    $\displaystyle
    \begin{gathered}
    \int {\frac{1}
    {{\left( {3 - 2x} \right)^2 }}dx} \hfill \\
    \hfill \\
    u = 3 - 2x \Rightarrow du = - 2dx \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \int {\frac{1}
    {{\left( {3 - 2x} \right)^2 }}dx} = \int {\frac{1}
    {{u^2 }} \cdot \frac{{ - 1}}
    {2}du} = \frac{{ - 1}}
    {2}\int {u^{ - 2} du} \Leftrightarrow \frac{{ - 1}}
    {2} \cdot \frac{{u^{ - 1} }}
    {{ - 1}} + C
    $

    $\displaystyle
    \therefore \boxed{\int {\frac{1}
    {{\left( {3 - 2x} \right)^2 }}dx} = \frac{1}
    {{2\left( {3 - 2x} \right)}} + C}
    $
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  5. #5
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    Nacho's answer is correct, but

    Quote Originally Posted by Nacho View Post

    $\displaystyle \frac{{ - 1}}
    {2}\int {u^{ - 2} du} \Leftrightarrow \frac{{ - 1}}
    {2} \cdot \frac{{u^{ - 1} }}
    {{ - 1}} + C
    $
    double implies it's applied when having equality, like $\displaystyle 2a=2b\iff a=b.$ So in this case we require an equal sign.
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