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Math Help - Problem in integration

  1. #1
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    Problem in integration

     \int \frac{x^3}{(x^2+5)^2} dx <br />

    <br />
u=x^2<br />
    <br />
du = 2xdx.\frac{x^2}{x^2}<br />
    <br />
\frac{udu}{2}=x^3dx<br />

     \frac{1}{2}\int \frac {udu}{(u+5)^2}

    \frac{1}{2}[\int \frac {du}{u+5} - \int \frac{5du}{(u+5)^2}]

    \frac{1}{2}[ln(u+5) +\frac{5}{(u+5)}]

    \frac{1}{2}[ln(x^2+5) +\frac{5}{(x^2+5)}]


    But the answer is different.

    Please help me finding which step is wrong.
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  2. #2
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    According to Wolfram you forgot the constant of integration: \frac{1}{2}[\ln(x^2+5) +\frac{5}{(x^2+5)}] + C

    The only real question I have is in the third line you've automatically assumed that x \neq 0 when multiplying by x^2/x^2
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  3. #3
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    \displaystyle \int{\frac{x^3}{(x^2 + 5)^2}\,dx} = \int{\frac{x^3}{x^4 + 10x^2 + 25}\,dx}

    \displaystyle = \frac{1}{4}\int{\frac{4x^3}{x^4 + 10x^2 + 25}\,dx}

    \displaystyle = \frac{1}{4}\int{\frac{4x^3 + 20x - 20x}{x^4 + 10x^2 + 25}\,dx}

    \displaystyle = \frac{1}{4}\int{\frac{4x^3 + 20x}{x^4 + 10x^2 + 25}\,dx} - \frac{1}{4}\int{\frac{20x}{x^4 + 10x^2 + 25}\,dx}

    \displaystyle = \frac{1}{4}\int{\frac{4x^3 + 20x}{x^4 + 10x^2 + 25}\,dx} - \frac{5}{2}\int{\frac{2x}{(x^2 + 5)^2}\,dx}.


    Now for the first make the substitution \displaystyle u = x^4 + 10x^2 + 25 and for the second make the substitution \displaystyle v= x^2 + 5.
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    According to Wolfram you forgot the constant of integration: \frac{1}{2}[\ln(x^2+5) +\frac{5}{(x^2+5)}] + C

    The only real question I have is in the third line you've automatically assumed that x \neq 0 when multiplying by x^2/x^2
    You mean to say that, the third step is wrong. What I have thought is, no matter what is the value of x, it can multipied and divided by a same value.

    But i am clueless to understand your point.

    Appreciate if you can help me.
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  5. #5
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    Quote Originally Posted by kumaran5555 View Post
    You mean to say that, the third step is wrong. What I have thought is, no matter what is the value of x, it can multipied and divided by a same value.

    But i am clueless to understand your point.

    Appreciate if you can help me.
    By that logic \frac{0}{0} = 1 which is surely undefined as it's division by 0?
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  6. #6
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    Thanks.

    So, I should not proceed with the approach I have used.

    Am I correct?
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  7. #7
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    Quote Originally Posted by kumaran5555 View Post
    Thanks.

    So, I should not proceed with the approach I have used.

    Am I correct?
    You should proceed with the approach I posted...
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  8. #8
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    Quote Originally Posted by Prove It View Post
    You should proceed with the approach I posted...
    I have got your approach.

    But my intention of this post is to understand what is wrong with my approach.

    Thanks.
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  9. #9
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    Quote Originally Posted by e^(i*pi) View Post
    By that logic \frac{0}{0} = 1 which is surely undefined as it's division by 0?
    As you said 0/0 is undefined, multiplying x , with x/x is not going to cause any undefined value.

    May be it's wrong multiplying a constant C with x/x and that may cause undefined results, when x=0.

    But in mycase  x.\frac {x^2}{x^2}
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  10. #10
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    It will still be undefined when x=0. It would be fine if 0 wasn't in the original domain but because it is you're still potentially dividing by 0. This means that x=0 is a potential solution. In your equation let f(x)= x \cdot \dfrac{x^2}{x^2} then f(0) = 0 \cdot \dfrac{0}{0}

    You can do it if you exclude 0 from the domain
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  11. #11
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    So, I am taking it as if any function which has zero in its domain, doing f(x).\frac{x}{x} will result in undefined value. So it should not be done.
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  12. #12
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    Yeah, that's it. Quite frustrating I admit
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