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Math Help - Stuck with Integration(Logarithms)

  1. #1
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    Stuck with Integration(Logarithms)

    This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

    \int \frac{x{dx}}{(x^2-1)ln(1-x^2)}

    \int\frac{sin4xdx}{1+sin^22x}

    \int\frac{\sqrt x dx}{1-\sqrt{x^3}}

    \int\frac{(2x-3) dx}{(x-2)^2}

    \int\frac{(3-4x)x dx}{\sqrt[3]{1-x}}
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    Senior Member DeMath's Avatar
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    Quote Originally Posted by honestliar View Post
    This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

    \int \frac{x{dx}}{(x^2-1)ln(1-x^2)}
    \int {\frac{x}{{\left( {{x^2} - 1} \right)\ln \left( {1 - {x^2}} \right)}}dx}  = \frac{1}{2}\int {\frac{{ - 2x}}{{\left( {1 - {x^2}} \right)\ln \left( {1 - {x^2}} \right)}}dx}  =

    = \left\{ \begin{gathered}\ln \left( {1 - {x^2}} \right) = t, \hfill \\\frac{{ - 2x}}{{1 - {x^2}}}dx = dt \hfill \\ \end{gathered}  \right\} = \frac{1}{2}\int {\frac{{dt}}{t}}  = \frac{1}{2}\ln \left| t \right| + C = \frac{1}<br />
{2}\ln \left| {\ln \left( {1 - {x^2}} \right)} \right| + C.
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    Senior Member DeMath's Avatar
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    Quote Originally Posted by honestliar View Post
    This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

    \int\frac{sin4xdx}{1+sin^22x}
    \int {\frac{{\sin 4x}}{{1 + {{\sin }^2}2x}}dx}  = \frac{1}{2}\int {\frac{{4\sin 2x\cos 2x}}{{1 + {{\sin }^2}2x}}dx}  = \left\{ \begin{gathered}1 + {\sin ^2}2x = t, \hfill \\4\sin 2x\cos 2xdx = dt \hfill \\ \end{gathered}  \right\} =

    = \frac{1}{2}\int {\frac{{dt}}{t}}  = \frac{1}{2}\ln \left| t \right| + C = \frac{1}{2}\ln \left( {1 + {{\sin }^2}2x} \right) + C.
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    Senior Member DeMath's Avatar
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    Quote Originally Posted by honestliar View Post
    This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

    \int\frac{\sqrt x dx}{1-\sqrt{x^3}}
    \int {\frac{{\sqrt x }}{{1 - \sqrt {{x^3}} }}dx}  = \int {\frac{{{x^{1/2}}}}{{1 - {x^{3/2}}}}dx}  =  - \frac{2}{3}\int {\frac{{ - \left( {3/2} \right){x^{1/2}}}}{{1 - {x^{3/2}}}}dx}  =

    = \left\{ \begin{gathered}1 - {x^{3/2}} = t, \hfill \\- \frac{3}<br />
{2}{x^{1/2}}dx = dt \hfill \\ \end{gathered}  \right\} =  - \frac{2}<br />
{3}\int {\frac{{dt}}{t}}  =  - \frac{2}{3}\ln \left| t \right| + C =  - \frac{2}{3}\ln \left| {1 - {x^{3/2}}} \right| + C.
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by honestliar View Post
    This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

    \int\frac{(2x-3) dx}{(x-2)^2}
    \int {\frac{{2x - 3}}{{{{\left( {x - 2} \right)}^2}}}dx}  = \int {\frac{{2\left( {x - 2} \right) + 1}}{{{{\left( {x - 2} \right)}^2}}}dx}  = 2\int {\frac{{dx}}{{x - 2}}}  + \int {\frac{{dx}}{{{{\left( {x - 2} \right)}^2}}}}  =2 \ln \left| {x - 2} \right| - \frac{1}{{x - 2}} + C.
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  6. #6
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    For the last one, use the sub: u = 1-x \ \Rightarrow \ -du = dx

    To get: -\int \frac{\left[3-4(1-u)\right]\left(1-u\right) du}{u^{\frac{1}{3}}} \ = \ \cdots \ = \ \int \frac{4u^2 - 5u + 1}{u^{\frac{1}{3}}}du

    Divide through using exponent laws and it'll be a breeze.
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