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Math Help - 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
    <br />
\int_0^1 {\frac{{2x + 3}}{{(x + 1)^2 }}} dx<br />

    my attempt

    <br />
\begin{array}{l}<br />
 \frac{{2x + 3}}{{(x + 1)(x + 1)}} = \frac{A}{{x + 1}} + \frac{B}{{(x + 1)^2 }} \\ <br />
 \frac{{2x + 3}}{{(x + 1)^2 }} = A(x + 1) + B \\ <br />
 2x + 3 = A(x) + (A + B) \\ <br />
 \end{array}<br />

    giving coefficients as

    <br />
\begin{array}{l}<br />
 A = 2 \\ <br />
 A + B = 3 \\ <br />
 so \\ <br />
 B = 1 \\ <br />
 \end{array}<br />

    and so the original integral is equal to

    <br />
\int_0^1 {2\frac{1}{{x + 1}}dx}  + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} <br />

    <br />
 = 2\left[ {\ln |x + 1|} \right]_0^1  + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} <br />

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

    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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    \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; March 17th 2009 at 03:36 AM. Reason: ..
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  3. #3
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    thanks

    <br />
\begin{array}{l}<br />
 \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx}  = u^{ - 2} dx \\ <br />
 u = x + 1 \\ <br />
 du = 1dx \\ <br />
 \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx}  = u^{ - 2} du \\ <br />
 \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  \\ <br />
 \end{array}<br />

    and so

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

    <br />
\begin{array}{l}<br />
 \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx}  = u^{ - 2} dx \\ <br />
 u = x + 1 \\ <br />
 du = 1dx \\ <br />
 \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx}  = u^{ - 2} du \\ <br />
 \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  \\ <br />
 \end{array}<br />

    and so

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