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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 1 hour 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









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      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



























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