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Math Help - Another Integration Problem - Understanding Numerator

  1. #1
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    Another Integration Problem - Understanding Numerator

    \int \dfrac{x + 3}{x^{2} + 2x + 2}dx

    \dfrac{1}{2} \int \dfrac{(x + 1) 2dx}{(x + 1)^{2} + 1} + \int \dfrac{2dx}{(x + 1)^{2} + 1} - What's going on with the numerator? I understand the denominator (completing the square).

    For instance, how can dx be fixed (u substitution)? Given: u = x^{2} + 2x + 2 and du = 2x + 2. Note: du and dx have to be fixed before splitting the fraction into two fractions.

    \dfrac{1}{2} \ln(x^{2} + 2x + 2) + 2 \arctan(x + 1) + C - Final answer
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    Re: Another Integration Problem - Understanding Numerator

    Jason
    It seems to me that you lack basic knowledge about integration.
    I suggest you download asap a calculus book and start to revise the chapter on integration.
    you may go here : Pauls Online Math Notes

    Now some hints...if u=x^2+2x+2 then du =(2x+2)dx therefore dx=(du)/(2x+2). du is the differential of the function u and as such it is equal to the derivative of u times the diferrential of x which is dx...
    There is nothing wrong with the numerator....the first integral can be manipulated to become ( the numerator) d[(x+1)^2+1] and as such it will give the ln(x^2+2x+2) after integration. the second one is related to the inverse trigonometric functions...do you have any idea of what I am talking about? please revise the related topic.
    Anyway the answer you post is correct.
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  3. #3
    Senior Member x3bnm's Avatar
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    Re: Another Integration Problem - Understanding Numerator

    \begin{align*}&\int \frac{x + 3}{x^{2} + 2x + 2}dx\\ =& \int \left( \frac{2}{x^2 + 2x + 2} + \frac{x+1}{x^{2} + 2x + 2}\right)\,\,dx...\text{[by algebra]} \\ =& \int \frac{2}{x^2 + 2x + 2}\,\, dx + \int \frac{x+1}{x^{2} + 2x + 2}\,\,dx....\text{[by property of integration]} \\=& 2 \int \frac{1}{(x + 1)^2 + 1^2}\,\,dx + \frac{1}{2} \int \frac{2(x+1)}{x^{2} + 2x + 2}\,\,dx\\  \\& \text{Let } u = x+1 \,\,\therefore du = dx\\& \text{And Let } m = x^{2} + 2x + 2\,\, \therefore dm = 2(x+1)\,\,dx\\\\=& 2\int \frac{1}{u^2 + 1^2}\,\,du + \frac{1}{2}\int \frac{1}{m}\,\,dm....\text{[by plugging in } m, dm, u, du\text{ ]}\\=& 2\tan^{-1}(u) + \frac{1}{2}\ln|m| + C\\=& 2\tan^{-1}(x+1) + \frac{1}{2}\ln|x^2 + 2x + 2| + C....\text{[by substituting } u \text{ and } m\text{ ]}\end{align*}

    Hope it helps.
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