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Math Help - intergration with partial fractions

  1. #1
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    intergration with partial fractions

    I am doing this problem:
    integral of (x^3 + x - 1)/((X^2 +1))^2 dx
    So i set that integral equal to (Ax + B)/(x^2 + 1) + (Cx + D)/((x^2 + 1))^2
    I got ax^3 + ax + bx^2 + b + cx + d = 1x^3 + 1x - 1
    so..
    x^3 terms: a = 1
    x^2 terms: b = 0
    x^1 terms: a + c = 1 ( c = 0)
    x^0 terms: b + d = -1 ( d = -1)
    I put these back into the equation and got
    integral of 1/(x^2 + 1) + integral of -1/((x^2 + 1))^2
    which equals ln(x^2 + 1) + 1/(x^2 + 1) + C
    The answer in the book has those 2 first terms as the answer..but has 2 more terms, then the Constant..I am having trouble finding the correct answer. What did I do wrong?
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  2. #2
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    First

    \int {\frac{{x^3 + x - 1}}<br />
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{{x\left( {x^2 + 1} \right) - 1}}<br />
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{x}<br />
{{\left( {x^2 + 1} \right)}}~dx} - \int {\frac{1}<br />
{{\left( {x^2 + 1} \right)^2 }}~dx}

    The second integral is already a partial fraction, apply integration by parts.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    First

    \int {\frac{{x^3 + x - 1}}<br />
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{{x\left( {x^2 + 1} \right) - 1}}<br />
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{x}<br />
{{\left( {x^2 + 1} \right)}}~dx} - \int {\frac{1}<br />
{{\left( {x^2 + 1} \right)^2 }}~dx}

    The second integral is already a partial fraction, apply integration by parts.
    wait..what did i do wrong? I am confused on what you are telling me.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    I am doing this problem:
    integral of (x^3 + x - 1)/((X^2 +1))^2 dx
    So i set that integral equal to (Ax + B)/(x^2 + 1) + (Cx + D)/((x^2 + 1))^2
    I got ax^3 + ax + bx^2 + b + cx + d = 1x^3 + 1x - 1
    so..
    x^3 terms: a = 1
    x^2 terms: b = 0
    x^1 terms: a + c = 1 ( c = 0)
    x^0 terms: b + d = -1 ( d = -1)
    I put these back into the equation and got
    integral of 1/(x^2 + 1) + integral of -1/((x^2 + 1))^2
    which equals ln(x^2 + 1) + 1/(x^2 + 1) + C
    The answer in the book has those 2 first terms as the answer..but has 2 more terms, then the Constant..I am having trouble finding the correct answer. What did I do wrong?
    Quote Originally Posted by Krizalid View Post
    First

    \int {\frac{{x^3 + x - 1}}<br />
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{{x\left( {x^2 + 1} \right) - 1}}<br />
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{x}<br />
{{\left( {x^2 + 1} \right)}}~dx} - \int {\frac{1}<br />
{{\left( {x^2 + 1} \right)^2 }}~dx}

    The second integral is already a partial fraction, apply integration by parts.
    Quote Originally Posted by davecs77 View Post
    wait..what did i do wrong? I am confused on what you are telling me.
    Both you are Krizalid are coming up with the same integrals. So look to the second integral for possible problems. For reference, my calculator gives:
    \frac{ln(x^2 + 1)}{2} - \frac{1}{2}tan^{-1}(x) - \frac{x}{2(x^2 + 1)} + C

    -Dan
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