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Math Help - Integral

  1. #1
    Member srirahulan's Avatar
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    Lightbulb Integral

    find out the value of the integration by using partial fraction knowledge, \int_{0}^{k}\frac{2}{(x+1)^2(x^2+1)}d(x) where [k>0]then \lim_{k\rightarrow\propto}\int_{0}^{k}\frac{2}{(x+  1)^2(x^2+1)}d(x),,,,I go through this way,,, first of all i find the value of the integral, it comes like this, \frac{1}{2}ln\frac{(k+1)^2}{k^2+1}+\frac{k}{k+1}then can i change intho this and find out the limit for that,, \frac{1}{2}ln\frac{(\frac{1}{k}+1)^2}{\frac{1}{k^2  }+1}+\frac{1}{\frac{1}{k}+1} Is It Correct or Not???
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  2. #2
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    Re: Integral

    Hello, srirahulan!

    Evaulate the integral with partial fractions: . \int_{0}^{k}\frac{2\,dx}{(x+1)^2(x^2+1)} where k>0
    then find: / \lim_{k\to\infty}\int_{0}^{k}\frac{2\,dx}{(x+1)^2(  x^2+1)}

    We have: . \frac{2}{(x+1)^2(x^2+1)} \;=\;\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{Cx}{x^2+1} + \frac{D}{x^2+1}

    . . 2 \;=\;A(x+1)(x^2+1) + B(x^2+1) + Cx(x+1`)^2 D(x+1)^2


    \text{Let }x = \text{-}1\!:\;2 \;=\;A(0) + B(2) + C(0) + D(0) \quad\Rightarrow\quad \boxed{B = 1}\;\;[1]

    \text{Let }x = 0\!:\;2 \;=\;A(1) + B(1) + C(0) + D(1)\;\;[2]

    \text{Let }x = 1\!:\;2 \;=\;A(4) + B(2) + C(4) + D(4)\;\;[3]

    \text{Let }x = 2\!:\;2 \;=\;A(15) + B(5) + C(18) + D(9)\;\;[4]


    \begin{array}{cccccccc}\text{[2] becomes:} & A + 1 + D &=&2 & \Rightarrow & A+D &=& 1 \\ \text{[3] becomes:} & 4A+ 2 + 4C + 4D &=& 2 & \Rightarrow & A+C+D &=& 0 \\ \text{[4] becomes:} & 15A + 5 + 18C + 9D &=& 2 & \Rightarrow & 5A + 6B + 3B &=& \text{-}1 \end{array}

    \text{Solve the system: }\:\begin{Bmatrix}A&=&1 \\ B&=&1 \\ C&=&\text{-}1 \\ D&=&0 \end{Bmatrix}

    \text{Hence: }\:\int^k_0\frac{2\,dx}{(x+1)^2(x^2+1)} \;=\;\int^k_0\left[\frac{1}{x+1} + \frac{1}{(x+1)^2} - \frac{x}{x^2+1}\right]\,dx

    . . . =\;\ln(x+1) - \frac{1}{x+1} - \tfrac{1}{2}\ln|x^2+1|\,\bigg]^k_0

    . . . =\;\tfrac{1}{2}\left[2\ln(x+1) - \frac{2}{1+x} - \ln(x^2+1)\right]^k_0

    . . . =\;\tfrac{1}{2}\left[\ln\frac{(x+1)^2}{x^2+1} = \frac{2}{x+1}\right]^k_0


    Can you continue?
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  3. #3
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    Re: Integral

    both of you please read up on the new LaTex interpreter.
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  4. #4
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    Re: Integral

    Quote Originally Posted by Soroban View Post
    Hello, srirahulan!


    We have: .$$\frac{2}{(x+1)^2(x^2+1)} \;=\;\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{Cx}{x^2+1} + \frac{D}{x^2+1}$$

    . . $$2 \;=\;A(x+1)(x^2+1) + B(x^2+1) + Cx(x+1`)^2 D(x+1)^2$$


    $$\text{Let }x = \text{-}1\!:\;2 \;=\;A(0) + B(2) + C(0) + D(0) \quad\Rightarrow\quad \boxed{B = 1}\;\;[1]$$

    $$\text{Let }x = 0\!:\;2 \;=\;A(1) + B(1) + C(0) + D(1)\;\;[2]$$

    $$\text{Let }x = 1\!:\;2 \;=\;A(4) + B(2) + C(4) + D(4)\;\;[3]$$

    $$\text{Let }x = 2\!:\;2 \;=\;A(15) + B(5) + C(18) + D(9)\;\;[4] $$


    $$\begin{array}{cccccccc}\text{[2] becomes:} & A + 1 + D &=&2 & \Rightarrow & A+D &=& 1 \\ \text{[3] becomes:} & 4A+ 2 + 4C + 4D &=& 2 & \Rightarrow & A+C+D &=& 0 \\ \text{[4] becomes:} & 15A + 5 + 18C + 9D &=& 2 & \Rightarrow & 5A + 6B + 3B &=& \text{-}1 \end{array}$$

    $$\text{Solve the system: }\:\begin{Bmatrix}A&=&1 \\ B&=&1 \\ C&=&\text{-}1 \\ D&=&0 \end{Bmatrix}$$

    $$\text{Hence: }\:\int^k_0\frac{2\,dx}{(x+1)^2(x^2+1)} \;=\;\int^k_0\left[\frac{1}{x+1} + \frac{1}{(x+1)^2} - \frac{x}{x^2+1}\right]\,dx$$

    . . . $$=\;\ln(x+1) - \frac{1}{x+1} - \tfrac{1}{2}\ln|x^2+1|\,\bigg]^k_0$$

    . . . $$=\;\tfrac{1}{2}\left[2\ln(x+1) - \frac{2}{1+x} - \ln(x^2+1)\right]^k_0$$

    . . . $$=\;\tfrac{1}{2}\left[\ln\frac{(x+1)^2}{x^2+1} = \frac{2}{x+1}\right]^k_0$$


    Can you continue?
    redone for new latex
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  5. #5
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    Re: Integral

    Quote Originally Posted by srirahulan View Post
    find out the value of the integration by using partial fraction knowledge,$$\int_{0}^{k}\frac{2}{(x+1)^2(x^2+1)}d( x)$$ where [k>0]then $$\lim_{k\rightarrow\propto}\int_{0}^{k}\frac{2}{( x+1)^2(x^2+1)}d(x)$$,,,,I go through this way,,, first of all i find the value of the integral, it comes like this, $$\frac{1}{2}ln\frac{(k+1)^2}{k^2+1}+\frac{k}{k+1} $$then can i change intho this and find out the limit for that,,$$\frac{1}{2}ln\frac{(\frac{1}{k}+1)^2}{ \frac{1}{k^2}+1}+\frac{1}{\frac{1}{k}+1}$$ Is It Correct or Not???
    redone for new LaTex
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  6. #6
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    Re: Integral

    OP, yes, you can change it just as you suggested.
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  7. #7
    Member srirahulan's Avatar
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    Re: Integral

    okay i got that, Thanks for everybody>>>>>>
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