How to calculate $int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx$ [closed]












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How to calculate $$int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} spacemathrm{d}x$$






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closed as off-topic by RRL, amWhy, A. Pongrácz, Gibbs, egreg Jan 6 at 22:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, A. Pongrácz, Gibbs, egreg

If this question can be reworded to fit the rules in the help center, please edit the question.





















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    How to calculate $$int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} spacemathrm{d}x$$






    definite-integrals







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    $endgroup$



    closed as off-topic by RRL, amWhy, A. Pongrácz, Gibbs, egreg Jan 6 at 22:53


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, A. Pongrácz, Gibbs, egreg

    If this question can be reworded to fit the rules in the help center, please edit the question.



















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      How to calculate $$int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} spacemathrm{d}x$$






      definite-integrals







      share|cite|improve this question











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      How to calculate $$int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} spacemathrm{d}x$$






      definite-integrals




      definite-integrals






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      share|cite|improve this question













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      edited Jan 6 at 20:52









      amWhy

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      asked Jan 6 at 17:25









      Adil AndersonAdil Anderson

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      closed as off-topic by RRL, amWhy, A. Pongrácz, Gibbs, egreg Jan 6 at 22:53


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, A. Pongrácz, Gibbs, egreg

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by RRL, amWhy, A. Pongrácz, Gibbs, egreg Jan 6 at 22:53


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, A. Pongrácz, Gibbs, egreg

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
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          Short answer: $I=1$.




          Proof:
          Note that
          begin{align*}
          I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
          overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
          overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
          end{align*}



          Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$






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            $$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
            So
            $mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.



            Then $2mathbf I =int_2^4 mathrm{dx}=2$.






            share|cite|improve this answer











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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$


              Short answer: $I=1$.




              Proof:
              Note that
              begin{align*}
              I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
              overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
              overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
              end{align*}



              Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$


                Short answer: $I=1$.




                Proof:
                Note that
                begin{align*}
                I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
                overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
                overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
                end{align*}



                Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$


                  Short answer: $I=1$.




                  Proof:
                  Note that
                  begin{align*}
                  I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
                  overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
                  overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
                  end{align*}



                  Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$






                  share|cite|improve this answer











                  $endgroup$




                  Short answer: $I=1$.




                  Proof:
                  Note that
                  begin{align*}
                  I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
                  overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
                  overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
                  end{align*}



                  Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 6 at 17:55









                  Bernard

                  119k740113




                  119k740113










                  answered Jan 6 at 17:28









                  Maximilian JanischMaximilian Janisch

                  44110




                  44110























                      1












                      $begingroup$

                      $$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
                      So
                      $mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.



                      Then $2mathbf I =int_2^4 mathrm{dx}=2$.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        $$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
                        So
                        $mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.



                        Then $2mathbf I =int_2^4 mathrm{dx}=2$.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          $$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
                          So
                          $mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.



                          Then $2mathbf I =int_2^4 mathrm{dx}=2$.






                          share|cite|improve this answer











                          $endgroup$



                          $$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
                          So
                          $mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.



                          Then $2mathbf I =int_2^4 mathrm{dx}=2$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 6 at 17:56









                          Bernard

                          119k740113




                          119k740113










                          answered Jan 6 at 17:33









                          Thomas ShelbyThomas Shelby

                          2,550421




                          2,550421















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