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Thread: integration of (2x+3)/(x+1)^2 for 0 to 1 by partial fractions

  1. #1
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    integration of (2x+3)/(x+1)^2 for 0 to 1 by partial fractions

    Question is to evaluate by partial fractions
    $\displaystyle
    \int_0^1 {\frac{{2x + 3}}{{(x + 1)^2 }}} dx
    $

    my attempt

    $\displaystyle
    \begin{array}{l}
    \frac{{2x + 3}}{{(x + 1)(x + 1)}} = \frac{A}{{x + 1}} + \frac{B}{{(x + 1)^2 }} \\
    \frac{{2x + 3}}{{(x + 1)^2 }} = A(x + 1) + B \\
    2x + 3 = A(x) + (A + B) \\
    \end{array}
    $

    giving coefficients as

    $\displaystyle
    \begin{array}{l}
    A = 2 \\
    A + B = 3 \\
    so \\
    B = 1 \\
    \end{array}
    $

    and so the original integral is equal to

    $\displaystyle
    \int_0^1 {2\frac{1}{{x + 1}}dx} + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx}
    $

    $\displaystyle
    = 2\left[ {\ln |x + 1|} \right]_0^1 + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx}
    $

    I think i'm tripping up here on the 2nd integral
    I had
    $\displaystyle
    \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = \left[ {\frac{1}{2}\ln (x + 1)} \right]_0^1
    $

    But I don't think this is right. Can someone show me how to integrate the 2nd integral? Thanks.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    $\displaystyle \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = \left[ {\frac{-1}{x+1}} \right]_0^1$

    Try substituting x+1 = u
    dx= du
    So basically you need to integrate 1/u^2
    -after integrating substitute u by x+1
    -then put the limits
    Last edited by ADARSH; Mar 17th 2009 at 02:36 AM. Reason: ..
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  3. #3
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    thanks

    $\displaystyle
    \begin{array}{l}
    \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = u^{ - 2} dx \\
    u = x + 1 \\
    du = 1dx \\
    \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = u^{ - 2} du \\
    \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = \left[ {\frac{{u^{ - 1} }}{{ - 1}}} \right]_0^1 = \left[ {\frac{{ - 1}}{{x + 1}}} \right]_0^1 \\
    \end{array}
    $

    and so

    $\displaystyle
    \begin{array}{l}
    \int_0^1 {2\frac{1}{{x + 1}}dx} + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} \\
    = 2\left[ {\ln |x + 1|} \right]_0^1 + \left[ {\frac{{ - 1}}{{x + 1}}} \right]_0^1 \\
    = 2\ln 2 + \frac{1}{2} \\
    \end{array}
    $
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Craka View Post
    thanks

    $\displaystyle
    \begin{array}{l}
    \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = u^{ - 2} dx \\
    u = x + 1 \\
    du = 1dx \\
    \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = u^{ - 2} du \\
    \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = \left[ {\frac{{u^{ - 1} }}{{ - 1}}} \right]_0^1 = \left[ {\frac{{ - 1}}{{x + 1}}} \right]_0^1 \\
    \end{array}
    $

    and so

    $\displaystyle
    \begin{array}{l}
    \int_0^1 {2\frac{1}{{x + 1}}dx} + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} \\
    = 2\left[ {\ln |x + 1|} \right]_0^1 + \left[ {\frac{{ - 1}}{{x + 1}}} \right]_0^1 \\
    = 2\ln 2 + \frac{1}{2} \\
    \end{array}
    $
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