Shortcut for value of this indefinite integral?How can this indefinite integral be solved without partial fractions?Question about indefinite integral with square rootIndefinite Integral of $n$-th power of Quadratic DenominatorIndefinite integral of a rational function problem…Need help in indefinite integral eliminationWeird indefinite integral situationHow to find the value of this indefinite integral?Indefinite integral of $arctan(x)$, why consider $1cdot dx$?Indefinite integral with polynomial function factorizingHow do I evaluate this indefinite integral?

How does Loki do this?

Pre-amplifier input protection

Is there a problem with hiding "forgot password" until it's needed?

Pole-zeros of a real-valued causal FIR system

Can the discrete variable be a negative number?

Return the Closest Prime Number

Is expanding the research of a group into machine learning as a PhD student risky?

How do I go from 300 unfinished/half written blog posts, to published posts?

What is the opposite of 'gravitas'?

How to check is there any negative term in a large list?

Would this custom Sorcerer variant that can only learn any verbal-component-only spell be unbalanced?

How to run a prison with the smallest amount of guards?

Is there a good way to store credentials outside of a password manager?

Was Spock the First Vulcan in Starfleet?

Where does the Z80 processor start executing from?

What does 算不上 mean in 算不上太美好的日子?

What does "I’d sit this one out, Cap," imply or mean in the context?

Is exact Kanji stroke length important?

Shortcut for value of this indefinite integral?

Increase performance creating Mandelbrot set in python

Did Dumbledore lie to Harry about how long he had James Potter's invisibility cloak when he was examining it? If so, why?

Opposite of a diet

How to safely derail a train during transit?

Different result between scanning in Epson's "color negative film" mode and scanning in positive -> invert curve in post?



Shortcut for value of this indefinite integral?


How can this indefinite integral be solved without partial fractions?Question about indefinite integral with square rootIndefinite Integral of $n$-th power of Quadratic DenominatorIndefinite integral of a rational function problem…Need help in indefinite integral eliminationWeird indefinite integral situationHow to find the value of this indefinite integral?Indefinite integral of $arctan(x)$, why consider $1cdot dx$?Indefinite integral with polynomial function factorizingHow do I evaluate this indefinite integral?













3












$begingroup$


If $$f(x) = int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx$$ 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)). $










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 $$f(x) = int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx$$ 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)). $










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 $$f(x) = int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx$$ 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)). $










share|cite|improve this question











$endgroup$




If $$f(x) = int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx$$ 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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 59 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


















5












$begingroup$

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 $fracepi4 - frace2 +1$.






share|cite|improve this answer









$endgroup$




















    4












    $begingroup$

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



    Now for $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx $$, do the following manipulation:
    $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx =int e^x biggr(arctan x - frac 11+x^2+frac 11+x^2+frac 2x(1+x^2)^2biggr),dx. $$



    Note that $$biggr(arctan x - frac 11+x^2biggr)'=frac 11+x^2+frac 2x(1+x^2)^2. $$



    Then by the above formula $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx=e^x biggr(arctan x - frac 11+x^2biggr)+c.$$
    So $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$






    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









      5












      $begingroup$

      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 $fracepi4 - frace2 +1$.






      share|cite|improve this answer









      $endgroup$

















        5












        $begingroup$

        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 $fracepi4 - frace2 +1$.






        share|cite|improve this answer









        $endgroup$















          5












          5








          5





          $begingroup$

          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 $fracepi4 - frace2 +1$.






          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 $fracepi4 - frace2 +1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Catalin ZaraCatalin Zara

          3,817514




          3,817514





















              4












              $begingroup$

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



              Now for $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx $$, do the following manipulation:
              $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx =int e^x biggr(arctan x - frac 11+x^2+frac 11+x^2+frac 2x(1+x^2)^2biggr),dx. $$



              Note that $$biggr(arctan x - frac 11+x^2biggr)'=frac 11+x^2+frac 2x(1+x^2)^2. $$



              Then by the above formula $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx=e^x biggr(arctan x - frac 11+x^2biggr)+c.$$
              So $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$






              share|cite|improve this answer









              $endgroup$

















                4












                $begingroup$

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



                Now for $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx $$, do the following manipulation:
                $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx =int e^x biggr(arctan x - frac 11+x^2+frac 11+x^2+frac 2x(1+x^2)^2biggr),dx. $$



                Note that $$biggr(arctan x - frac 11+x^2biggr)'=frac 11+x^2+frac 2x(1+x^2)^2. $$



                Then by the above formula $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx=e^x biggr(arctan x - frac 11+x^2biggr)+c.$$
                So $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$






                share|cite|improve this answer









                $endgroup$















                  4












                  4








                  4





                  $begingroup$

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



                  Now for $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx $$, do the following manipulation:
                  $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx =int e^x biggr(arctan x - frac 11+x^2+frac 11+x^2+frac 2x(1+x^2)^2biggr),dx. $$



                  Note that $$biggr(arctan x - frac 11+x^2biggr)'=frac 11+x^2+frac 2x(1+x^2)^2. $$



                  Then by the above formula $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx=e^x biggr(arctan x - frac 11+x^2biggr)+c.$$
                  So $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$






                  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 $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx $$, do the following manipulation:
                  $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx =int e^x biggr(arctan x - frac 11+x^2+frac 11+x^2+frac 2x(1+x^2)^2biggr),dx. $$



                  Note that $$biggr(arctan x - frac 11+x^2biggr)'=frac 11+x^2+frac 2x(1+x^2)^2. $$



                  Then by the above formula $$int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx=e^x biggr(arctan x - frac 11+x^2biggr)+c.$$
                  So $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Thomas ShelbyThomas Shelby

                  4,4992726




                  4,4992726



























                      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

                      How should I use the fbox command correctly to avoid producing a Bad Box message?How to put a long piece of text in a box?How to specify height and width of fboxIs there an arrayrulecolor-like command to change the rule color of fbox?What is the command to highlight bad boxes in pdf?Why does fbox sometimes place the box *over* the graphic image?how to put the text in the boxHow to create command for a box where text inside the box can automatically adjust?how can I make an fbox like command with certain color, shape and width of border?how to use fbox in align modeFbox increase the spacing between the box and it content (inner margin)how to change the box height of an equationWhat is the use of the hbox in a newcommand command?

                      Doxepinum Nexus interni Notae | Tabula navigationis3158DB01142WHOa682390"Structural Analysis of the Histamine H1 Receptor""Transdermal and Topical Drug Administration in the Treatment of Pain""Antidepressants as antipruritic agents: A review"

                      inputenc: Unicode character … not set up for use with LaTeX The Next CEO of Stack OverflowEntering Unicode characters in LaTeXHow to solve the `Package inputenc Error: Unicode char not set up for use with LaTeX` problem?solve “Unicode char is not set up for use with LaTeX” without special handling of every new interesting UTF-8 characterPackage inputenc Error: Unicode character ² (U+B2)(inputenc) not set up for use with LaTeX. acroI2C[I²C]package inputenc error unicode char (u + 190) not set up for use with latexPackage inputenc Error: Unicode char u8:′ not set up for use with LaTeX. 3′inputenc Error: Unicode char u8: not set up for use with LaTeX with G-BriefPackage Inputenc Error: Unicode char u8: not set up for use with LaTeXPackage inputenc Error: Unicode char ́ (U+301)(inputenc) not set up for use with LaTeX. includePackage inputenc Error: Unicode char ̂ (U+302)(inputenc) not set up for use with LaTeX. … $widehatleft (OA,AA' right )$Package inputenc Error: Unicode char â„¡ (U+2121)(inputenc) not set up for use with LaTeX. printbibliography[heading=bibintoc]Package inputenc Error: Unicode char − (U+2212)(inputenc) not set up for use with LaTeXPackage inputenc Error: Unicode character α (U+3B1) not set up for use with LaTeXPackage inputenc Error: Unicode characterError: ! Package inputenc Error: Unicode char ⊘ (U+2298)(inputenc) not set up for use with LaTeX