Stuck at indefinite integral












4












$begingroup$


I'm completely stuck on solving this indefinite integral:
$$intfrac{x-2}{-x^2+2x-5}dx$$



By completing the square in the denominator and separating the original into two integrals, I get:



$$-intfrac{x}{x^2-2x+5}dx -intfrac{2}{(x-1)^2 + 4}dx$$



The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.






integration indefinite-integrals partial-fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
    $endgroup$
    – J.G.
    3 hours ago










  • $begingroup$
    Let $u$ be the denominator and notice how $du$ nicely falls out.
    $endgroup$
    – John Douma
    3 hours ago










  • $begingroup$
    @JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfrac{xdu}{(2x-2)u}$?
    $endgroup$
    – Arcturus
    3 hours ago












  • $begingroup$
    You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
    $endgroup$
    – Larry
    3 hours ago
















4












$begingroup$


I'm completely stuck on solving this indefinite integral:
$$intfrac{x-2}{-x^2+2x-5}dx$$



By completing the square in the denominator and separating the original into two integrals, I get:



$$-intfrac{x}{x^2-2x+5}dx -intfrac{2}{(x-1)^2 + 4}dx$$



The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.






integration indefinite-integrals partial-fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
    $endgroup$
    – J.G.
    3 hours ago










  • $begingroup$
    Let $u$ be the denominator and notice how $du$ nicely falls out.
    $endgroup$
    – John Douma
    3 hours ago










  • $begingroup$
    @JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfrac{xdu}{(2x-2)u}$?
    $endgroup$
    – Arcturus
    3 hours ago












  • $begingroup$
    You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
    $endgroup$
    – Larry
    3 hours ago














4












4








4





$begingroup$


I'm completely stuck on solving this indefinite integral:
$$intfrac{x-2}{-x^2+2x-5}dx$$



By completing the square in the denominator and separating the original into two integrals, I get:



$$-intfrac{x}{x^2-2x+5}dx -intfrac{2}{(x-1)^2 + 4}dx$$



The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.






integration indefinite-integrals partial-fractions







share|cite|improve this question











$endgroup$




I'm completely stuck on solving this indefinite integral:
$$intfrac{x-2}{-x^2+2x-5}dx$$



By completing the square in the denominator and separating the original into two integrals, I get:



$$-intfrac{x}{x^2-2x+5}dx -intfrac{2}{(x-1)^2 + 4}dx$$



The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.






integration indefinite-integrals partial-fractions




integration indefinite-integrals partial-fractions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 mins ago









DavidG

1,988620




1,988620










asked 3 hours ago









ArcturusArcturus

695




695












  • $begingroup$
    Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
    $endgroup$
    – J.G.
    3 hours ago










  • $begingroup$
    Let $u$ be the denominator and notice how $du$ nicely falls out.
    $endgroup$
    – John Douma
    3 hours ago










  • $begingroup$
    @JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfrac{xdu}{(2x-2)u}$?
    $endgroup$
    – Arcturus
    3 hours ago












  • $begingroup$
    You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
    $endgroup$
    – Larry
    3 hours ago


















  • $begingroup$
    Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
    $endgroup$
    – J.G.
    3 hours ago










  • $begingroup$
    Let $u$ be the denominator and notice how $du$ nicely falls out.
    $endgroup$
    – John Douma
    3 hours ago










  • $begingroup$
    @JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfrac{xdu}{(2x-2)u}$?
    $endgroup$
    – Arcturus
    3 hours ago












  • $begingroup$
    You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
    $endgroup$
    – Larry
    3 hours ago
















$begingroup$
Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
$endgroup$
– J.G.
3 hours ago




$begingroup$
Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
$endgroup$
– J.G.
3 hours ago












$begingroup$
Let $u$ be the denominator and notice how $du$ nicely falls out.
$endgroup$
– John Douma
3 hours ago




$begingroup$
Let $u$ be the denominator and notice how $du$ nicely falls out.
$endgroup$
– John Douma
3 hours ago












$begingroup$
@JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfrac{xdu}{(2x-2)u}$?
$endgroup$
– Arcturus
3 hours ago






$begingroup$
@JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfrac{xdu}{(2x-2)u}$?
$endgroup$
– Arcturus
3 hours ago














$begingroup$
You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
$endgroup$
– Larry
3 hours ago




$begingroup$
You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
$endgroup$
– Larry
3 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):



begin{align}intfrac{x-2}{-x^2+2x-5}dx &= - int frac{x-2}{x^2 - 2x + 5}dx
\&= -frac 12 int frac{2x-4}{x^2 - 2x + 5}dx. end{align} The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields



begin{align}
intfrac{x-2}{-x^2+2x-5}dx = -frac 12 left(int frac{2x-2}{x^2-2x+5}dx - int frac{2}{x^2-2x+5}dxright).
end{align}



The second integral you can solve, can you solve the first?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
    $endgroup$
    – Arcturus
    2 hours ago



















4












$begingroup$

Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
begin{align*}I &= -int frac{x-2}{(x-1)^2+4},dx\
&phantom{|}^{u=x-1}_{du=dx}\
&= int -frac{u-1}{u^2+4},du = intfrac{-u}{u^2+4},du+intfrac{1}{u^2+4},duend{align*}
The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $$I=intfrac{x-2}{-x^2+2x-5}dx$$
    $$I=-intfrac{x-2}{x^2-2x+5}dx$$
    $$I=-frac12intfrac{2x-2-2}{x^2-2x+5}dx$$
    $$I=-frac12intfrac{2x-2}{x^2-2x+5}dx+intfrac{dx}{x^2-2x+5}$$
    $$I=-frac12I_1+I_2$$




    $$I_1=intfrac{2x-2}{x^2-2x+5}dx$$
    $u=x^2-2x+5Rightarrow du=(2x-2)dx$:
    $$I_1=intfrac{du}u$$
    $$I_1=ln|u|$$
    $$I_1=ln|x^2-2x+5|$$
    $$I_1=ln(x^2-2x+5)$$




    $$I_2=text{something you know how to solve}$$




    $$I=-frac12ln(x^2-2x+5)+I_2$$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):



      begin{align}intfrac{x-2}{-x^2+2x-5}dx &= - int frac{x-2}{x^2 - 2x + 5}dx
      \&= -frac 12 int frac{2x-4}{x^2 - 2x + 5}dx. end{align}       The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields



      begin{align}
      intfrac{x-2}{-x^2+2x-5}dx = -frac 12 left(int frac{2x-2}{x^2-2x+5}dx - int frac{2}{x^2-2x+5}dxright).
      end{align}



      The second integral you can solve, can you solve the first?






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
        $endgroup$
        – Arcturus
        2 hours ago
















      4












      $begingroup$

      The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):



      begin{align}intfrac{x-2}{-x^2+2x-5}dx &= - int frac{x-2}{x^2 - 2x + 5}dx
      \&= -frac 12 int frac{2x-4}{x^2 - 2x + 5}dx. end{align}       The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields



      begin{align}
      intfrac{x-2}{-x^2+2x-5}dx = -frac 12 left(int frac{2x-2}{x^2-2x+5}dx - int frac{2}{x^2-2x+5}dxright).
      end{align}



      The second integral you can solve, can you solve the first?






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
        $endgroup$
        – Arcturus
        2 hours ago














      4












      4








      4





      $begingroup$

      The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):



      begin{align}intfrac{x-2}{-x^2+2x-5}dx &= - int frac{x-2}{x^2 - 2x + 5}dx
      \&= -frac 12 int frac{2x-4}{x^2 - 2x + 5}dx. end{align}       The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields



      begin{align}
      intfrac{x-2}{-x^2+2x-5}dx = -frac 12 left(int frac{2x-2}{x^2-2x+5}dx - int frac{2}{x^2-2x+5}dxright).
      end{align}



      The second integral you can solve, can you solve the first?






      share|cite|improve this answer











      $endgroup$



      The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):



      begin{align}intfrac{x-2}{-x^2+2x-5}dx &= - int frac{x-2}{x^2 - 2x + 5}dx
      \&= -frac 12 int frac{2x-4}{x^2 - 2x + 5}dx. end{align}       The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields



      begin{align}
      intfrac{x-2}{-x^2+2x-5}dx = -frac 12 left(int frac{2x-2}{x^2-2x+5}dx - int frac{2}{x^2-2x+5}dxright).
      end{align}



      The second integral you can solve, can you solve the first?







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 3 hours ago

























      answered 3 hours ago









      E-muE-mu

      620314




      620314








      • 1




        $begingroup$
        Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
        $endgroup$
        – Arcturus
        2 hours ago














      • 1




        $begingroup$
        Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
        $endgroup$
        – Arcturus
        2 hours ago








      1




      1




      $begingroup$
      Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
      $endgroup$
      – Arcturus
      2 hours ago




      $begingroup$
      Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
      $endgroup$
      – Arcturus
      2 hours ago











      4












      $begingroup$

      Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
      begin{align*}I &= -int frac{x-2}{(x-1)^2+4},dx\
      &phantom{|}^{u=x-1}_{du=dx}\
      &= int -frac{u-1}{u^2+4},du = intfrac{-u}{u^2+4},du+intfrac{1}{u^2+4},duend{align*}
      The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
        begin{align*}I &= -int frac{x-2}{(x-1)^2+4},dx\
        &phantom{|}^{u=x-1}_{du=dx}\
        &= int -frac{u-1}{u^2+4},du = intfrac{-u}{u^2+4},du+intfrac{1}{u^2+4},duend{align*}
        The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
          begin{align*}I &= -int frac{x-2}{(x-1)^2+4},dx\
          &phantom{|}^{u=x-1}_{du=dx}\
          &= int -frac{u-1}{u^2+4},du = intfrac{-u}{u^2+4},du+intfrac{1}{u^2+4},duend{align*}
          The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.






          share|cite|improve this answer









          $endgroup$



          Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
          begin{align*}I &= -int frac{x-2}{(x-1)^2+4},dx\
          &phantom{|}^{u=x-1}_{du=dx}\
          &= int -frac{u-1}{u^2+4},du = intfrac{-u}{u^2+4},du+intfrac{1}{u^2+4},duend{align*}
          The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          jmerryjmerry

          3,409413




          3,409413























              1












              $begingroup$

              $$I=intfrac{x-2}{-x^2+2x-5}dx$$
              $$I=-intfrac{x-2}{x^2-2x+5}dx$$
              $$I=-frac12intfrac{2x-2-2}{x^2-2x+5}dx$$
              $$I=-frac12intfrac{2x-2}{x^2-2x+5}dx+intfrac{dx}{x^2-2x+5}$$
              $$I=-frac12I_1+I_2$$




              $$I_1=intfrac{2x-2}{x^2-2x+5}dx$$
              $u=x^2-2x+5Rightarrow du=(2x-2)dx$:
              $$I_1=intfrac{du}u$$
              $$I_1=ln|u|$$
              $$I_1=ln|x^2-2x+5|$$
              $$I_1=ln(x^2-2x+5)$$




              $$I_2=text{something you know how to solve}$$




              $$I=-frac12ln(x^2-2x+5)+I_2$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $$I=intfrac{x-2}{-x^2+2x-5}dx$$
                $$I=-intfrac{x-2}{x^2-2x+5}dx$$
                $$I=-frac12intfrac{2x-2-2}{x^2-2x+5}dx$$
                $$I=-frac12intfrac{2x-2}{x^2-2x+5}dx+intfrac{dx}{x^2-2x+5}$$
                $$I=-frac12I_1+I_2$$




                $$I_1=intfrac{2x-2}{x^2-2x+5}dx$$
                $u=x^2-2x+5Rightarrow du=(2x-2)dx$:
                $$I_1=intfrac{du}u$$
                $$I_1=ln|u|$$
                $$I_1=ln|x^2-2x+5|$$
                $$I_1=ln(x^2-2x+5)$$




                $$I_2=text{something you know how to solve}$$




                $$I=-frac12ln(x^2-2x+5)+I_2$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $$I=intfrac{x-2}{-x^2+2x-5}dx$$
                  $$I=-intfrac{x-2}{x^2-2x+5}dx$$
                  $$I=-frac12intfrac{2x-2-2}{x^2-2x+5}dx$$
                  $$I=-frac12intfrac{2x-2}{x^2-2x+5}dx+intfrac{dx}{x^2-2x+5}$$
                  $$I=-frac12I_1+I_2$$




                  $$I_1=intfrac{2x-2}{x^2-2x+5}dx$$
                  $u=x^2-2x+5Rightarrow du=(2x-2)dx$:
                  $$I_1=intfrac{du}u$$
                  $$I_1=ln|u|$$
                  $$I_1=ln|x^2-2x+5|$$
                  $$I_1=ln(x^2-2x+5)$$




                  $$I_2=text{something you know how to solve}$$




                  $$I=-frac12ln(x^2-2x+5)+I_2$$






                  share|cite|improve this answer









                  $endgroup$



                  $$I=intfrac{x-2}{-x^2+2x-5}dx$$
                  $$I=-intfrac{x-2}{x^2-2x+5}dx$$
                  $$I=-frac12intfrac{2x-2-2}{x^2-2x+5}dx$$
                  $$I=-frac12intfrac{2x-2}{x^2-2x+5}dx+intfrac{dx}{x^2-2x+5}$$
                  $$I=-frac12I_1+I_2$$




                  $$I_1=intfrac{2x-2}{x^2-2x+5}dx$$
                  $u=x^2-2x+5Rightarrow du=(2x-2)dx$:
                  $$I_1=intfrac{du}u$$
                  $$I_1=ln|u|$$
                  $$I_1=ln|x^2-2x+5|$$
                  $$I_1=ln(x^2-2x+5)$$




                  $$I_2=text{something you know how to solve}$$




                  $$I=-frac12ln(x^2-2x+5)+I_2$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  clathratusclathratus

                  3,541332




                  3,541332






























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