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Thread: Confused with Integration Problem

  1. #1
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    Confused with Integration Problem

    Can you please show me how to solve this integral? I = (x+2)/(x^2+x) dx
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  2. #2
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    Re: Confused with Integration Problem

    Have you tried partial fraction decomposition?
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  3. #3
    Senior Member x3bnm's Avatar
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    Re: Confused with Integration Problem

    By partial fraction expansion:

    $\displaystyle \begin{align*}\frac{A}{x} + \frac{B}{x+1} =& \frac{x+2}{x^2+x}......\text{(1)}\\ A(x+1) + Bx =& x + 2\\ (A+B)x + A =& x + 2\end{align*}$

    $\displaystyle A + B = 1......\text{(2)}$

    $\displaystyle A = 2.............\text{(3)}$

    $\displaystyle \therefore\,\,B = -1$

    Now plugging these values of $\displaystyle A$ and $\displaystyle B$ into (1) and integrating both sides:

    $\displaystyle \begin{align*}\int \frac{x+2}{x^2+x}\,\,dx =& \int \left(\frac{A}{x} + \frac{B}{x+1}\right)\,\,dx\\ =& \int \frac{2}{x}\,\,dx - \int \frac{1}{x+1}\,\,dx\\ =& 2\ln{(x)} - \ln{(x+1)} + C\end{align*}$
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  4. #4
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    Re: Confused with Integration Problem

    Quote Originally Posted by x3bnm View Post
    By partial fraction expansion:

    $\displaystyle \begin{align*}\frac{A}{x} + \frac{B}{x+1} =& \frac{x+2}{x^2+x}......\text{(1)}\\ A(x+1) + Bx =& x + 2\\ (A+B)x + A =& x + 2\end{align*}$

    $\displaystyle A + B = 1......\text{(2)}$

    $\displaystyle A = 2.............\text{(3)}$

    $\displaystyle \therefore\,\,B = -1$

    Now plugging these values of $\displaystyle A$ and $\displaystyle B$ into (1) and integrating both sides:

    $\displaystyle \begin{align*}\int \frac{x+2}{x^2+x}\,\,dx =& \int \left(\frac{A}{x} + \frac{B}{x+1}\right)\,\,dx\\ =& \int \frac{2}{x}\,\,dx - \int \frac{1}{x+1}\,\,dx\\ =& 2\ln{(x)} - \ln{(x+1)} + C\end{align*}$
    Actually it's $\displaystyle \displaystyle \begin{align*} 2\ln{|x|} - \ln{|x+1|} + C \end{align*}$.
    Thanks from jdesak
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