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Math Help - Rational integral

  1. #1
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    Rational integral

    Solve \displaystyle\int\frac{2x^3+3x^2+x-1}{(x+1)(x^2+2x+2)^2}\,dx by not using partial fractions.

    I liked this particular integral, I solved it on another forum to show that sometimes we can avoid the annoying algebra when doing the partial fractions.
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  2. #2
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     \displaystyle   \int \frac{2x^3 + 3x^2 + x - 1 }{(x+1)(x^2 + 2x+2)^2} ~dx

      \displaystyle  =\int \frac{(x+1)(2x^2 + x ) - 1 }{(x+1)(x^2 + 2x+2)^2} ~dx

      \displaystyle  =  \int \frac{(x+1)(2x^2 + 2x+1 ) - ((x+1)^2 + 1) }{(x+1)(x^2 + 2x+2)^2} ~dx

      \displaystyle  = \int \frac{ 2x^2 + 2x + 1 }{ (x^2 + 2x + 2 )^2 } ~dx - \int \frac{dx}{(x+1)(x^2 + 2x + 2 ) }

      \displaystyle  = \int \frac{ (x^2 + 2x + 2 ) + (x^2 - 1) }{ (x^2 + 2x + 2 )^2 } ~dx - \int \frac{dx}{(x+1)(x^2 + 2x + 2 ) }

      \displaystyle  = \int \frac{ dx}{ x^2 + 2x + 2  }   + \int \frac{(x-1)(x+1)}{(x+1)(x^2 + 2x + 2 )^2 } ~dx - \int \frac{dx}{(x+1)(x^2 + 2x + 2 ) }

    Sub.   \displaystyle  x+1 = 1/t on the third integral , we have

      \displaystyle  \int \frac{dx}{(x+1)(x^2 + 2x + 2 ) }  = - \int \frac{t~dt}{t^2 + 1 }

     = - \frac{1}{2} \ln(t^2+ 1 ) + C = \ln|1+x| - \frac{1}{2} \ln(x^2 + 2x + 2) + C

    For the second integral , use integration by parts ,

      \displaystyle  \int \frac{(x-1)(x+1)}{ (x^2 + 2x + 2 )^2 } ~dx

      \displaystyle  = - \frac{1}{2} ~ \frac{x-1}{x^2+2x+2} +  \frac{1}{2}~\int \frac{dx}{x^2 + 2x + 2  }


    Therefore ,

      \displaystyle  \int \frac{2x^3 + 3x^2 + x - 1 }{(x+1)(x^2 + 2x+2)^2} ~dx

      \displaystyle  = \frac{3}{2}~ \int \frac{ dx}{ x^2 + 2x + 2  } - \frac{1}{2} ~ \frac{x-1}{x^2+2x+2}  - \ln|1+x| + \frac{1}{2} \ln(x^2 + 2x + 2) + C'

      \displaystyle  = \frac{3}{2} \tan^{-1}(x+1)   - \frac{1}{2} ~ \frac{x-1}{x^2+2x+2}  - \ln|1+x| + \frac{1}{2} \ln(x^2 + 2x + 2) + C'
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  3. #3
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    I observe that:

    2x^3+3x^2+x-1 = (x+1)(x^2+2x+2)-(x^2+2x+2)+(x-1) (x+1)^2

    So we have:

    \begin{aligned} \displaystyle\int\frac{2x^3+3x^2+x-1}{(x+1)(x^2+2x+2)^2}\;dx  &=\int\frac{(x+1)(x^2+2x+2)-(x^2+2x+2)+(x-1) (x+1)^2}{(x+1)(x^2+2x+2)^2}\;dx  \\ & = \int \frac{1}{x^2+2x+2}-\frac{1}{(x+1)(x^2+2x+2)}+\frac{x^2-1}{(x^2+2x+2)^2}\;{dx} \\ & = \int \frac{1}{(x+1)^2+1} \;{dx}-\int\frac{(x^2+2x+2)-(x+1)^2}{(x+1)(x^2+2x+2)}\;{dx}+\int \frac{x^2-1}{(x^2+2x+2)^2}\;{dx} \\ & = \arctan(x+1)-\int\frac{1}{x+1}\;{dx}+\int\frac{(x+1)}{(x+1)^2+1  }\;{dx}+\int \frac{x^2-1}{(x^2+2x+2)^2}\;{dx} \\ &  = \arctan(x+1)-\ln(x+1)+\frac{1}{2}\int\frac{2x+2}{x^2+2x+2}\;{dx  }+\int \frac{x^2-1}{(x^2+2x+2)^2}\;{dx}}  \\ & = \arctan(x+1)-\ln(x+1)+\frac{1}{2}\ln(x^2+2x+2)+\int \frac{x^2-1}{(x^2+2x+2)^2}\;{dx}  \\ & \end{aligned}

    Where for the last integral we use the substitution t = (x+1) to get:

    \begin{aligned} \int \frac{x^2-1}{(x^2+2x+2)^2}\;{dx} & = \int\frac{(x+1)(x-1)}{[(x+1)^2+1]^2}\;{dx} = \int\frac{t^2-2t}{(t^2+1)^2}\;{dt}\\& = \int\frac{t^2}{(t^2+1)^2}\;{dt}-\int\frac{2t}{(t^2+1)^2}\;{dt}\\& = \frac{1}{t^2+1}+\int\frac{t^2}{(t^2+1)^2}\;{dt}\\& \end{aligned}

    Letting t = \tan{u} in the remaining integral we have:


    \begin{aligned} \int\frac{t^2}{(t^2+1)^2}\;{dt} &  = \int\frac{\tan^2{u}\sec^2{u}}{\sec^4{u}}\;{du} = \int \sin^2{u}\;{du} =  \frac{1}{2}\;{u}-\frac{1}{4}\sin{2u}+k \\& = \frac{1}{2}\arctan{t}-\frac{1}{4}\sin\left(2\arctan{t}\right)+k = \frac{1}{2}\arctan{t}-\frac{t}{2(t^2+1)}+k \\& = \frac{1}{2}\arctan(x+1)-\frac{x+1}{2(x^2+2x+2)}+k \end{aligned}

    Thus


    \displaystyle\int\frac{2x^3+3x^2+x-1}{(x+1)(x^2+2x+2)^2}\;dx  \displaystyle  = \frac{3}{2}\arctan(x+1)+\ln\left|\frac{1}{x+1}\rig  ht|+\frac{1}{2}\ln\left|(x+1)^2+1\right|  \displaystyle -\frac{1}{2}\left(\frac{x+1}{(x+1)^2+1}\right)+\lef  t(\frac{1}{(x+1)^2+1}\right)+k
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  4. #4
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    good! pretty different solutions from the one i have:

    consider \displaystyle\int{\frac{{{t}^{2}}}{{{\left( {{t}^{2}}+1 \right)}^{2}}}\,dt}=\int{t\left( -\frac{1}{2\left( {{t}^{2}}+1 \right)} \right)'\,dt}=-\frac{t}{2\left( {{t}^{2}}+1 \right)}+\frac12\int{\frac{dt}{{{t}^{2}}+1}}, (1); now from \displaystyle\int\frac{dt}{t(t^2+1)^2} put u=\dfrac1{t^2}+1 and the integral becomes \displaystyle-\frac12\int\frac{u-1}{u^2}\,du, (2).

    on the original integral put t=x+1 then

    \begin{aligned}<br />
   \int{\frac{2{{x}^{3}}+3{{x}^{2}}+x-1}{(x+1){{\left( {{x}^{2}}+2x+2 \right)}^{2}}}\,dx}&=\int{\frac{2{{x}^{3}}+3{{x}^{  2}}+x-1}{(x+1){{\left( {{(x+1)}^{2}}+1 \right)}^{2}}}\,dx} \\ <br />
 & =\int{\frac{2{{t}^{3}}-3{{t}^{2}}+t-1}{t{{\left( {{t}^{2}}+1 \right)}^{2}}}\,dt} \\ <br />
 & =\int{\frac{{{t}^{2}}}{{{\left( {{t}^{2}}+1 \right)}^{2}}}\,dt}-3\int{\frac{t}{{{\left( {{t}^{2}}+1 \right)}^{2}}}\,dt}+\int{\frac{dt}{{{t}^{2}}+1}}-\int{\frac{dt}{t{{\left( {{t}^{2}}+1 \right)}^{2}}}},<br />
\end{aligned}

    from (1) and (2) the latter equals

    \displaystyle-\frac{t}{2\left( {{t}^{2}}+1 \right)}+\frac32\arctan (t)+\frac{3}{2\left( {{t}^{2}}+1 \right)}+\frac{1}{2}\left[ \ln \left(\frac{{{t}^{2}}+1}{{{t}^{2}}}\right)+\frac{{  {t}^{2}}}{{{t}^{2}}+1} \right]+k,

    and the original integral equals

    \displaystyle-\frac{x+1}{2\left( {{(x+1)}^{2}}+1 \right)}+\frac{3}{2}\arctan (x+1)+\frac{3}{2\left( {{(x+1)}^{2}}+1 \right)}+\frac{1}{2}\left[ \ln \left( \frac{{{(x+1)}^{2}}+1}{{{(x+1)}^{2}}} \right)+\frac{{{(x+1)}^{2}}}{{{(x+1)}^{2}}+1} \right]+k.
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  5. #5
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    What about letting x + 1 = \tan \theta right away. The integral becomes

    \displaystyle \int \dfrac{2 \tan^3 \theta - 3 \tan^2 \theta + \tan \theta - 1}{\tan \theta \sec^2 \theta} d \theta

    or

    \displaystyle \int 2 \sin^2 \theta - 3 \sin \theta \cos \theta + \cos^2 \theta - \frac{\cos^3 \theta }{\sin \theta }\; d \theta

    or

    \displaystyle \int \frac{3}{2} - \frac{1}{2} \cos 2 \theta - 2 \sin \theta \cos \theta - \dfrac{\cos \theta}{\sin \theta} \; d \theta

    Integrating gives

    \frac{3}{2} \theta - \frac{1}{4} \sin 2 \theta - \sin^2 \theta - \ln \sin \theta  + c

    Then backsubstitute noting that

    \theta = \tan^{-1}(x+1),\;\; \sin \theta = \dfrac{x+1}{\sqrt{x^2+2x+2}},\;\; \cos \theta = \dfrac{1}{\sqrt{x^2+2x+2}}.
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