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Math Help - What did I do wrong on this integral?

  1. #1
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    What did I do wrong on this integral?

    Hi,

    I have an integral $\int \frac{6x^3}{x^3+1}$. First thing I do is long division to get $\int 6 - \int \frac{6}{x^3+1}$. To solve the second integral, I expand the denominator using the sum of cubes rule and then use partial fraction decomposition and find that A = 2, B = 2, and C = 4. This gives me the integrals $\int \frac{2}{x+1} + \int \frac{2x+4}{x^2-x+1}$. Solving the first one is easy, it's just $2 \ln (x+1)$. The second integral I think I solved correctly by first adding 5 and -5 to get $\int \frac{2x+4 - 5}{x^2-x+1} + \int \frac{5}{x^2-x+1}$. Now I can use u substitution on the first integral to get $\ln (x^2-x+1)$ and on the second integral I complete the square: $\int \frac{5}{(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}$. Solving this integral gives me $\frac{2}{\sqrt{3}}tan^{-1} \left( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)$.

    Final answer is:

    $2\ln (x+1) + \ln (x^2-x+1) + \frac{2}{\sqrt{3}}tan^{-1} \left( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + 6x$

    The problem is that when I enter this into my online assignment I'm told I'm wrong but I can't see my error. Does anyone see it?

    Thanks.
    Last edited by KevinShaughnessy; April 28th 2014 at 08:02 PM.
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  2. #2
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    Re: What did I do wrong on this integral?

    I'm seeing a bunch of disagreement in constants, what are most likely algebra errors.
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  3. #3
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    Re: What did I do wrong on this integral?

    Quote Originally Posted by KevinShaughnessy View Post
    Hi,

    I have an integral $\int \frac{6x^3}{x^3+1}$. First thing I do is long division to get $\int 6 - \int \frac{6}{x^3+1}$. To solve the second integral, I expand the denominator using the sum of cubes rule and then use partial fraction decomposition and find that A = 2, B = 2, and C = 4. This gives me the integrals $\int \frac{2}{x+1} + \int \frac{2x+4}{x^2-x+1}$. Solving the first one is easy, it's just $2 \ln (x+1)$. The second integral I think I solved correctly by first adding 5 and -5 to get $\int \frac{2x+4 - 5}{x^2-x+1} + \int \frac{5}{x^2-x+1}$. Now I can use u substitution on the first integral to get $\ln (x^2-x+1)$ and on the second integral I complete the square: $\int \frac{5}{(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}$. Solving this integral gives me $\frac{2}{\sqrt{3}}tan^{-1} \left( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)$.

    Final answer is:

    $2\ln (x+1) + \ln (x^2-x+1) + \frac{2}{\sqrt{3}}tan^{-1} \left( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + 6x$

    The problem is that when I enter this into my online assignment I'm told I'm wrong but I can't see my error. Does anyone see it?

    Thanks.
    $\displaystyle \begin{align*} \frac{A}{x + 1} + \frac{B\,x + C}{x^2 - x + 1} &\equiv \frac{6}{x^3 + 1} \\ A \left( x^2 - x + 1 \right) + \left( B\,x + C \right) \left( x + 1 \right) &\equiv 6 \\ A\,x^2 - A\,x + A + B\,x^2 + B\,x + C\,x + C &\equiv 6 \\ \left( A + B \right) x^2 + \left( B + C - A \right) x + \left( A + C \right) &\equiv 0x^2 + 0x + 6 \end{align*}$

    giving the first equation

    $\displaystyle \begin{align*} A+ B = 0 \implies A = -B \end{align*}$

    substituting into the second

    $\displaystyle \begin{align*} B + C - A = 0 \implies B + C - (-B) = 0 \implies 2B + C = 0 \implies C = -2B \end{align*}$.

    Substituting into the third

    $\displaystyle \begin{align*} A + C = 6 \implies -B - 2B = 6 \implies -3B = 6 \implies B = -2 \end{align*}$

    and so $\displaystyle \begin{align*} A = 2 \end{align*}$ and $\displaystyle \begin{align*} C = 4 \end{align*}$

    so your partial fraction decomposition for $\displaystyle \begin{align*} \frac{6}{x^3 + 1} \end{align*}$ is actually $\displaystyle \begin{align*} \frac{2}{x + 1} + \frac{-2x + 4}{x^2 - x + 1} \end{align*}$.
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