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Thread: Major Integration Help

  1. #1
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    Major Integration Help

    A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

    1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

    2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

    3. int [ tan 5x * dx ]
    ___An even problem, is the answer:
    ___-ln | csc(2x) + cot(2x) | + c

    4. int [ sec (x/2) * dx ]
    ___An even problem, is the answer:
    ___2*ln | sec(x/2) + tan(x/2) | + c

    5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

    Thanks Again!
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  2. #2
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    Quote Originally Posted by dillonmhudson View Post
    A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

    1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

    2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

    3. int [ tan 5x * dx ]
    ___An even problem, is the answer:
    ___-ln | csc(2x) + cot(2x) | + c

    4. int [ sec (x/2) * dx ]
    ___An even problem, is the answer:
    ___2*ln | sec(x/2) + tan(x/2) | + c

    5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

    Thanks Again!
    1. Are you sure the numerator isn't $\displaystyle x^2 + 3x + 2$???
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  3. #3
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    Quote Originally Posted by dillonmhudson View Post
    A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

    1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

    2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

    3. int [ tan 5x * dx ]
    ___An even problem, is the answer:
    ___-ln | csc(2x) + cot(2x) | + c

    4. int [ sec (x/2) * dx ]
    ___An even problem, is the answer:
    ___2*ln | sec(x/2) + tan(x/2) | + c

    5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

    Thanks Again!
    1. Long divide. Then the integrand becomes $\displaystyle x - 4 - \frac{2}{x + 1}$. Much easier.
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    You thinking about factoring? Yeah I'm positive, I double checked that.
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  5. #5
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    Quote Originally Posted by dillonmhudson View Post
    You thinking about factoring? Yeah I'm positive, I double checked that.
    Worth a shot.

    You'll have to long divide in that case.
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  6. #6
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    Quote Originally Posted by dillonmhudson View Post
    A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

    1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

    2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

    3. int [ tan 5x * dx ]
    ___An even problem, is the answer:
    ___-ln | csc(2x) + cot(2x) | + c

    4. int [ sec (x/2) * dx ]
    ___An even problem, is the answer:
    ___2*ln | sec(x/2) + tan(x/2) | + c

    5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

    Thanks Again!
    3. Remember that $\displaystyle \tan{5x} = \frac{\sin{5x}}{\cos{5x}}$.

    Make the substitution $\displaystyle u = \cos{5x}$.
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  7. #7
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    Quote Originally Posted by dillonmhudson View Post
    A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

    1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

    2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

    3. int [ tan 5x * dx ]
    ___An even problem, is the answer:
    ___-ln | csc(2x) + cot(2x) | + c

    4. int [ sec (x/2) * dx ]
    ___An even problem, is the answer:
    ___2*ln | sec(x/2) + tan(x/2) | + c

    5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

    Thanks Again!
    For 5. You'll have to long divide as well.

    $\displaystyle \frac{x^2 - 1}{x + 1} = x - 1 - \frac{1}{x + 1}$.
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  8. #8
    Senior Member DeMath's Avatar
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    Quote Originally Posted by dillonmhudson View Post
    A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

    1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

    2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

    3. int [ tan 5x * dx ]
    ___An even problem, is the answer:
    ___-ln | csc(2x) + cot(2x) | + c

    4. int [ sec (x/2) * dx ]
    ___An even problem, is the answer:
    ___2*ln | sec(x/2) + tan(x/2) | + c

    5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

    Thanks Again!
    $\displaystyle 1.{\text{ }}\int {\frac{{{x^2} - 3x + 2}}{{x + 1}}} dx = \int {\frac{{\left( {x + 1} \right)x - \left( {4x - 2} \right)}}{{x + 1}}} dx = \int {\left[ {x - \frac{{4x - 2}}{{x + 1}}} \right]} dx =$

    $\displaystyle = \frac{{{x^2}}}{2} - 2\int {\frac{{2x - 1}}{{x + 1}}} dx = \frac{{{x^2}}}{2} - 2\int {\frac{{2x + 2 - 3}}{{x + 1}}} dx = \frac{{{x^2}}}
    {2} - 2\int {\left[ {2 - \frac{3}{{x + 1}}} \right]} dx =$

    $\displaystyle = \frac{{{x^2}}}{2} - 4\int {dx} + 6\int {\frac{{dx}}{{x + 1}}} = \frac{{{x^2}}}{2} - 4x + 6\ln \left( {x + 1} \right) + C.$
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  9. #9
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    Quote Originally Posted by DeMath View Post

    $\displaystyle = \frac{{{x^2}}}{2} - 4\int {dx} + 6\int {\frac{{dx}}{{x + 1}}} = \frac{{{x^2}}}{2} - 4x + 6\ln \left( {x + 1} \right) + C.$
    Not quite on the $\displaystyle \ln(x+1),$ something is missing...
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  10. #10
    Senior Member DeMath's Avatar
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    This $\displaystyle \ln \left| {x + 1} \right|$?

    Thanks!
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  11. #11
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    Quote Originally Posted by Prove It View Post
    1. Long divide. Then the integrand becomes $\displaystyle x - 4 - \frac{2}{x + 1}$. Much easier.
    I think the integrand should be $\displaystyle x - 4 + \frac{6}{x + 1}$ after the long division.
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