Shortcut for value of this indefinite integral?












3












$begingroup$


If $$f(x) = int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx$$ and $f(0)=0$ then value of $f(1)$ is?



This is actually a Joint Entrance Examination question so I have to do it in two minutes. Is there a shortcut to find this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Actually the answer is $1 + e (pi/4 - 1/2)$. I would hate to have to do this in two minutes.
    $endgroup$
    – Robert Israel
    3 hours ago










  • $begingroup$
    @RobertIsrael. I was typing almost the same ! Cheers
    $endgroup$
    – Claude Leibovici
    3 hours ago










  • $begingroup$
    @RobertIsrael there must be a printing error in my book then.
    $endgroup$
    – Hema
    3 hours ago










  • $begingroup$
    What is JEE...?
    $endgroup$
    – amsmath
    3 hours ago










  • $begingroup$
    @amsmath Joint Entrance Exam in India. en.wikipedia.org/wiki/Joint_Entrance_Examination
    $endgroup$
    – Deepak
    2 hours ago
















3












$begingroup$


If $$f(x) = int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx$$ and $f(0)=0$ then value of $f(1)$ is?



This is actually a Joint Entrance Examination question so I have to do it in two minutes. Is there a shortcut to find this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Actually the answer is $1 + e (pi/4 - 1/2)$. I would hate to have to do this in two minutes.
    $endgroup$
    – Robert Israel
    3 hours ago










  • $begingroup$
    @RobertIsrael. I was typing almost the same ! Cheers
    $endgroup$
    – Claude Leibovici
    3 hours ago










  • $begingroup$
    @RobertIsrael there must be a printing error in my book then.
    $endgroup$
    – Hema
    3 hours ago










  • $begingroup$
    What is JEE...?
    $endgroup$
    – amsmath
    3 hours ago










  • $begingroup$
    @amsmath Joint Entrance Exam in India. en.wikipedia.org/wiki/Joint_Entrance_Examination
    $endgroup$
    – Deepak
    2 hours ago














3












3








3


1



$begingroup$


If $$f(x) = int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx$$ and $f(0)=0$ then value of $f(1)$ is?



This is actually a Joint Entrance Examination question so I have to do it in two minutes. Is there a shortcut to find this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$




If $$f(x) = int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx$$ and $f(0)=0$ then value of $f(1)$ is?



This is actually a Joint Entrance Examination question so I have to do it in two minutes. Is there a shortcut to find this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $






calculus integration indefinite-integrals




calculus integration indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 47 mins ago







Hema

















asked 3 hours ago









HemaHema

6531213




6531213








  • 2




    $begingroup$
    Actually the answer is $1 + e (pi/4 - 1/2)$. I would hate to have to do this in two minutes.
    $endgroup$
    – Robert Israel
    3 hours ago










  • $begingroup$
    @RobertIsrael. I was typing almost the same ! Cheers
    $endgroup$
    – Claude Leibovici
    3 hours ago










  • $begingroup$
    @RobertIsrael there must be a printing error in my book then.
    $endgroup$
    – Hema
    3 hours ago










  • $begingroup$
    What is JEE...?
    $endgroup$
    – amsmath
    3 hours ago










  • $begingroup$
    @amsmath Joint Entrance Exam in India. en.wikipedia.org/wiki/Joint_Entrance_Examination
    $endgroup$
    – Deepak
    2 hours ago














  • 2




    $begingroup$
    Actually the answer is $1 + e (pi/4 - 1/2)$. I would hate to have to do this in two minutes.
    $endgroup$
    – Robert Israel
    3 hours ago










  • $begingroup$
    @RobertIsrael. I was typing almost the same ! Cheers
    $endgroup$
    – Claude Leibovici
    3 hours ago










  • $begingroup$
    @RobertIsrael there must be a printing error in my book then.
    $endgroup$
    – Hema
    3 hours ago










  • $begingroup$
    What is JEE...?
    $endgroup$
    – amsmath
    3 hours ago










  • $begingroup$
    @amsmath Joint Entrance Exam in India. en.wikipedia.org/wiki/Joint_Entrance_Examination
    $endgroup$
    – Deepak
    2 hours ago








2




2




$begingroup$
Actually the answer is $1 + e (pi/4 - 1/2)$. I would hate to have to do this in two minutes.
$endgroup$
– Robert Israel
3 hours ago




$begingroup$
Actually the answer is $1 + e (pi/4 - 1/2)$. I would hate to have to do this in two minutes.
$endgroup$
– Robert Israel
3 hours ago












$begingroup$
@RobertIsrael. I was typing almost the same ! Cheers
$endgroup$
– Claude Leibovici
3 hours ago




$begingroup$
@RobertIsrael. I was typing almost the same ! Cheers
$endgroup$
– Claude Leibovici
3 hours ago












$begingroup$
@RobertIsrael there must be a printing error in my book then.
$endgroup$
– Hema
3 hours ago




$begingroup$
@RobertIsrael there must be a printing error in my book then.
$endgroup$
– Hema
3 hours ago












$begingroup$
What is JEE...?
$endgroup$
– amsmath
3 hours ago




$begingroup$
What is JEE...?
$endgroup$
– amsmath
3 hours ago












$begingroup$
@amsmath Joint Entrance Exam in India. en.wikipedia.org/wiki/Joint_Entrance_Examination
$endgroup$
– Deepak
2 hours ago




$begingroup$
@amsmath Joint Entrance Exam in India. en.wikipedia.org/wiki/Joint_Entrance_Examination
$endgroup$
– Deepak
2 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

With $g(t) = arctan(t) = tan^{-1}(t)$, the function is $$f(x) = int_0^x e^t (g(t) - g''(t)) , dt = int_0^x [e^t g(t)]' - [e^t g'(t)]', dt = $$ $$ = int_0^x [e^t(g(t) - g'(t)]' , dt =
e^x(g(x) - g'(x)) - (g(0) - g'(0))$$



As noted in comments, $f(1)$ is actually $frac{epi}{4} - frac{e}{2} +1$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Actually there is a formula $$int e^x (g (x)+g'(x)),dx = e^xcdot g (x)+c.$$



    Now for $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx $$, do the following manipulation:
    $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx =int e^x biggr(arctan x - frac {1}{1+x^2}+frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}biggr),dx. $$



    Note that $$biggr(arctan x - frac {1}{1+x^2}biggr)'=frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}. $$



    Then by the above formula $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx=e^x biggr(arctan x - frac {1}{1+x^2}biggr)+c.$$
    So $$f (1)=biggr[e^x biggr(arctan x - frac {1}{1+x^2}biggr)biggr]_0^1=frac {epi}{4}-frac {e}{2}+1. $$






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3165393%2fshortcut-for-value-of-this-indefinite-integral%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      With $g(t) = arctan(t) = tan^{-1}(t)$, the function is $$f(x) = int_0^x e^t (g(t) - g''(t)) , dt = int_0^x [e^t g(t)]' - [e^t g'(t)]', dt = $$ $$ = int_0^x [e^t(g(t) - g'(t)]' , dt =
      e^x(g(x) - g'(x)) - (g(0) - g'(0))$$



      As noted in comments, $f(1)$ is actually $frac{epi}{4} - frac{e}{2} +1$.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        With $g(t) = arctan(t) = tan^{-1}(t)$, the function is $$f(x) = int_0^x e^t (g(t) - g''(t)) , dt = int_0^x [e^t g(t)]' - [e^t g'(t)]', dt = $$ $$ = int_0^x [e^t(g(t) - g'(t)]' , dt =
        e^x(g(x) - g'(x)) - (g(0) - g'(0))$$



        As noted in comments, $f(1)$ is actually $frac{epi}{4} - frac{e}{2} +1$.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          With $g(t) = arctan(t) = tan^{-1}(t)$, the function is $$f(x) = int_0^x e^t (g(t) - g''(t)) , dt = int_0^x [e^t g(t)]' - [e^t g'(t)]', dt = $$ $$ = int_0^x [e^t(g(t) - g'(t)]' , dt =
          e^x(g(x) - g'(x)) - (g(0) - g'(0))$$



          As noted in comments, $f(1)$ is actually $frac{epi}{4} - frac{e}{2} +1$.






          share|cite|improve this answer









          $endgroup$



          With $g(t) = arctan(t) = tan^{-1}(t)$, the function is $$f(x) = int_0^x e^t (g(t) - g''(t)) , dt = int_0^x [e^t g(t)]' - [e^t g'(t)]', dt = $$ $$ = int_0^x [e^t(g(t) - g'(t)]' , dt =
          e^x(g(x) - g'(x)) - (g(0) - g'(0))$$



          As noted in comments, $f(1)$ is actually $frac{epi}{4} - frac{e}{2} +1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Catalin ZaraCatalin Zara

          3,807514




          3,807514























              3












              $begingroup$

              Actually there is a formula $$int e^x (g (x)+g'(x)),dx = e^xcdot g (x)+c.$$



              Now for $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx $$, do the following manipulation:
              $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx =int e^x biggr(arctan x - frac {1}{1+x^2}+frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}biggr),dx. $$



              Note that $$biggr(arctan x - frac {1}{1+x^2}biggr)'=frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}. $$



              Then by the above formula $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx=e^x biggr(arctan x - frac {1}{1+x^2}biggr)+c.$$
              So $$f (1)=biggr[e^x biggr(arctan x - frac {1}{1+x^2}biggr)biggr]_0^1=frac {epi}{4}-frac {e}{2}+1. $$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Actually there is a formula $$int e^x (g (x)+g'(x)),dx = e^xcdot g (x)+c.$$



                Now for $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx $$, do the following manipulation:
                $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx =int e^x biggr(arctan x - frac {1}{1+x^2}+frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}biggr),dx. $$



                Note that $$biggr(arctan x - frac {1}{1+x^2}biggr)'=frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}. $$



                Then by the above formula $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx=e^x biggr(arctan x - frac {1}{1+x^2}biggr)+c.$$
                So $$f (1)=biggr[e^x biggr(arctan x - frac {1}{1+x^2}biggr)biggr]_0^1=frac {epi}{4}-frac {e}{2}+1. $$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Actually there is a formula $$int e^x (g (x)+g'(x)),dx = e^xcdot g (x)+c.$$



                  Now for $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx $$, do the following manipulation:
                  $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx =int e^x biggr(arctan x - frac {1}{1+x^2}+frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}biggr),dx. $$



                  Note that $$biggr(arctan x - frac {1}{1+x^2}biggr)'=frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}. $$



                  Then by the above formula $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx=e^x biggr(arctan x - frac {1}{1+x^2}biggr)+c.$$
                  So $$f (1)=biggr[e^x biggr(arctan x - frac {1}{1+x^2}biggr)biggr]_0^1=frac {epi}{4}-frac {e}{2}+1. $$






                  share|cite|improve this answer









                  $endgroup$



                  Actually there is a formula $$int e^x (g (x)+g'(x)),dx = e^xcdot g (x)+c.$$



                  Now for $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx $$, do the following manipulation:
                  $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx =int e^x biggr(arctan x - frac {1}{1+x^2}+frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}biggr),dx. $$



                  Note that $$biggr(arctan x - frac {1}{1+x^2}biggr)'=frac {1}{1+x^2}+frac {2x}{(1+x^2)^2}. $$



                  Then by the above formula $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx=e^x biggr(arctan x - frac {1}{1+x^2}biggr)+c.$$
                  So $$f (1)=biggr[e^x biggr(arctan x - frac {1}{1+x^2}biggr)biggr]_0^1=frac {epi}{4}-frac {e}{2}+1. $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Thomas ShelbyThomas Shelby

                  4,4892726




                  4,4892726






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3165393%2fshortcut-for-value-of-this-indefinite-integral%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Ponta tanko

                      Image
                      Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
                      Read more

                      Tantalo (mitologio)

                      Image
                      Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
                      Read more

                      Erzsébet Schaár

                      Image
                      Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig...
                      Read more