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







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$endgroup$








  • 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 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






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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 $frac{epi}{4} - frac{e}{2} +1$.






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
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      active

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      2 Answers
      2






      active

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      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 $frac{epi}{4} - frac{e}{2} +1$.






      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 $frac{epi}{4} - frac{e}{2} +1$.






        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 $frac{epi}{4} - frac{e}{2} +1$.






          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

          3,817514




          3,817514























              4












              $begingroup$

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



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









                  Thomas ShelbyThomas Shelby

                  4,4992726




                  4,4992726






























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                      Tu-95轟炸機

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                      body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機...
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