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Math Help - long complicated integral.

  1. #1
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    long complicated integral.

    Does anyone know any shortcut to solve the following integral? As it stands the only way I know how to do it is to first use polynomial long division and then use integration with partial fractions. It is not necessarily difficult, it is just long and tedious, so I was wondering if there were any short cuts to lessen the chances of me making a careless error.

    \int\frac{2x^3+7x^2-15x+18}{2x^2+5x-3}
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  2. #2
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    Quote Originally Posted by gammaman View Post
    Does anyone know any shortcut to solve the following integral? As it stands the only way I know how to do it is to first use polynomial long division and then use integration with partial fractions. It is not necessarily difficult, it is just long and tedious, so I was wondering if there were any short cuts to lessen the chances of me making a careless error.

    \int\frac{2x^3+7x^2-15x+18}{2x^2+5x-3}
    do the long division and the partial fractions ... yes it's tedious, but you probably need the algebra practice anyway.
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  3. #3
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    <br />
\begin{gathered}<br />
  I = \int {\frac{{2x^3  + 7x^2  - 15x + 18}}<br />
{{2x^2  + 5x - 3}}dx = \int {\frac{{\left( {x + 1} \right)\left( {2x^2  + 5x - 3} \right) - 7x + 21}}<br />
{{2x^2  + 5x - 3}}dx} }  \hfill \\<br />
   \hfill \\<br />
  I = \int {\left( {x + 1} \right)dx - \underbrace {\int {\frac{{7x - 21}}<br />
{{2x^2  + 5x - 3}}dx} }_A}  \hfill \\ <br />
\end{gathered} <br />

    <br />
A = \int {\frac{{7x - 21}}<br />
{{2x^2  + 5x - 3}}dx}  = \frac{7}<br />
{4}\int {\frac{{4x - 12 + (5 - 5)}}<br />
{{2x^2  + 5x - 3}}dx}  = <br />
<br />
\frac{7}<br />
{4}\int {\frac{{4x - 12 + 5}}<br />
{{2x^2  + 5x - 3}}dx}  - \frac{7}<br />
{4}\int {\frac{{17}}<br />
{{2x^2  + 5x - 3}}dx} <br />

    <br />
A = \frac{7}<br />
{4}\int {\frac{{\left( {2x^2  + 5x - 3} \right)^\prime  }}<br />
{{2x^2  + 5x - 3}}dx}  - \frac{{7 \cdot 17}}<br />
{4}\int {\frac{1}<br />
{{\left( {2x + \frac{5}<br />
{2}} \right)^2  - \frac{{37}}<br />
{4}}}dx} <br />

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  4. #4
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    Quote Originally Posted by skeeter View Post
    do the long division and the partial fractions ... yes it's tedious, but you probably need the algebra practice anyway.
    Your right I could really use the algebra practice. The only thing I am not solid on is how to make sure that I factor

    2x^2+5x-3 correctly. I know that it is (2x-1)(x+3) but only because when you foil, you see that it is correct. Is there any short-cut to check that you factored correctly without having to foil?
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  5. #5
    Moo
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    Hello,
    Quote Originally Posted by Nacho View Post
    <br />
\begin{gathered}<br />
  I = \int {\frac{{2x^3  + 7x^2  - 15x + 18}}<br />
{{2x^2  + 5x - 3}}dx = \int {\frac{{\left( {x + 1} \right)\left( {2x^2  + 5x - 3} \right) - {\color{red}7}x + 21}}<br />
{{2x^2  + 5x - 3}}dx} }  \hfill \\<br />
\end{gathered} <br />
    I'm afraid it's -17x
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  6. #6
    Member Nacho's Avatar
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    Quote Originally Posted by gammaman View Post
    2x^2+5x-3 correctly. I know that it is (2x-1)(x+3) but only because when you foil, you see that it is correct. Is there any short-cut to check that you factored correctly without having to foil?
    <br />
2x^2  + 5x - 3 = \frac{{2\left( {2x^2  + 5x - 3} \right)}}<br />
{2} = \frac{{\left( {2x} \right)^2  + 5\left( {2x} \right) - 6}}<br />
{2} = \frac{{\left( {2x + 6} \right)\left( {2x - 1} \right)}}<br />
{2}<br />

    Quote Originally Posted by Moo View Post
    I'm afraid it's -17x
    Is true, thanks... but the importante is the idea
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