Shortcut for value of $f(1)$ where $f(x) = int e^x left(arctan x + frac {2x}{(1+x^2)^2}right),dx$












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 finding this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $






calculus integration indefinite-integrals







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  • 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
    2 days ago










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










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










  • $begingroup$
    What is JEE...?
    $endgroup$
    – amsmath
    2 days ago










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










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










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










  • $begingroup$
    What is JEE...?
    $endgroup$
    – amsmath
    2 days ago










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









YuiTo Cheng

2,1862937




2,1862937










asked 2 days ago









HemaHema

6621213




6621213








  • 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
    2 days ago










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










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










  • $begingroup$
    What is JEE...?
    $endgroup$
    – amsmath
    2 days ago










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










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










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










  • $begingroup$
    What is JEE...?
    $endgroup$
    – amsmath
    2 days ago










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












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




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












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




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












$begingroup$
What is JEE...?
$endgroup$
– amsmath
2 days ago




$begingroup$
What is JEE...?
$endgroup$
– amsmath
2 days ago












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




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










2 Answers
2






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6












$begingroup$

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












    $begingroup$

    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









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      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









      $endgroup$


















        6












        $begingroup$

        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









        $endgroup$
















          6












          6








          6





          $begingroup$

          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 2 days ago









          Catalin ZaraCatalin Zara

          3,827514




          3,827514























              4












              $begingroup$

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



              Now for $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx $$, 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









              $endgroup$


















                4












                $begingroup$

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



                Now for $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx $$, 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









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

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



                  Now for $$int e^x biggr(arctan x + frac {2x}{(1+x^2)^2}biggr),dx $$, 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 days ago









                  Thomas ShelbyThomas Shelby

                  4,5762727




                  4,5762727






























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                      南乌拉尔铁路局

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                      南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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