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

Shortcut/trick for integrating a factored polynomial?$int sqrtfracxx+1dx$Integrate $ int frac e^arctan(x)(1+x^2)^frac32 dx $Problem with definite integral $int _0^fracpi 2sin left(arctan left(xright)+xright)dx$$int frac2x9x^2+3dx=?$Integrate $int frac11+arctan(x)dx$A shortcut for solving $int fracdx(sin x+cos x)^3$Calculate $int_0^1fraccos(arctan x)sqrtxdx$About the integral $intarctanleft(frac1sinh^2 xright)dx$, some idea or feedbackFind $intarctan x,mathrm dx$ without substitution

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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 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

edited Mar 28 at 8:09

YuiTo Cheng

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asked Mar 28 at 2:28

HemaHema

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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
  Mar 28 at 2:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael. I was typing almost the same ! Cheers
  $endgroup$
  – Claude Leibovici
  Mar 28 at 2:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael there must be a printing error in my book then.
  $endgroup$
  – Hema
  Mar 28 at 2:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is JEE...?
  $endgroup$
  – amsmath
  Mar 28 at 2:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amsmath Joint Entrance Exam in India. en.wikipedia.org/wiki/Joint_Entrance_Examination
  $endgroup$
  – Deepak
  Mar 28 at 3:47
  

  
  
  
  

 | 
show 6 more comments

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 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

edited Mar 28 at 8:09

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Mar 28 at 2:28

HemaHema

67511 gold badge33 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Mar 28 at 2:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael. I was typing almost the same ! Cheers
  $endgroup$
  – Claude Leibovici
  Mar 28 at 2:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael there must be a printing error in my book then.
  $endgroup$
  – Hema
  Mar 28 at 2:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is JEE...?
  $endgroup$
  – amsmath
  Mar 28 at 2:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amsmath Joint Entrance Exam in India. en.wikipedia.org/wiki/Joint_Entrance_Examination
  $endgroup$
  – Deepak
  Mar 28 at 3:47
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

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 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

edited Mar 28 at 8:09

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Mar 28 at 2:28

HemaHema

67511 gold badge33 silver badges1414 bronze badges

$endgroup$

share|cite|improve this question

edited Mar 28 at 8:09

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Mar 28 at 2:28

HemaHema

67511 gold badge33 silver badges1414 bronze badges

share|cite|improve this question

edited Mar 28 at 8:09

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Mar 28 at 2:28

HemaHema

67511 gold badge33 silver badges1414 bronze badges

edited Mar 28 at 8:09

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

edited Mar 28 at 8:09

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Mar 28 at 2:28

HemaHema

67511 gold badge33 silver badges1414 bronze badges

asked Mar 28 at 2:28

HemaHema

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2 Answers
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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 $fracepi4 - frace2 +1$.

share|cite|improve this answer

answered Mar 28 at 2:51

Catalin ZaraCatalin Zara

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add a comment | 

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

answered Mar 28 at 3:43

Thomas ShelbyThomas Shelby

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add a comment | 

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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 $fracepi4 - frace2 +1$.

share|cite|improve this answer

answered Mar 28 at 2:51

Catalin ZaraCatalin Zara

4,03766 silver badges1515 bronze badges

$endgroup$

add a comment | 

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 $fracepi4 - frace2 +1$.

share|cite|improve this answer

answered Mar 28 at 2:51

Catalin ZaraCatalin Zara

4,03766 silver badges1515 bronze badges

$endgroup$

add a comment | 

$begingroup$

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

answered Mar 28 at 2:51

Catalin ZaraCatalin Zara

4,03766 silver badges1515 bronze badges

$endgroup$

share|cite|improve this answer

answered Mar 28 at 2:51

Catalin ZaraCatalin Zara

4,03766 silver badges1515 bronze badges

answered Mar 28 at 2:51

Catalin ZaraCatalin Zara

4,03766 silver badges1515 bronze badges

answered Mar 28 at 2:51

Catalin ZaraCatalin Zara

4,03766 silver badges1515 bronze badges

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

answered Mar 28 at 3:43

Thomas ShelbyThomas Shelby

6,39933 gold badges99 silver badges3232 bronze badges

$endgroup$

add a comment | 

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

answered Mar 28 at 3:43

Thomas ShelbyThomas Shelby

6,39933 gold badges99 silver badges3232 bronze badges

$endgroup$

add a comment | 

$begingroup$

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

answered Mar 28 at 3:43

Thomas ShelbyThomas Shelby

6,39933 gold badges99 silver badges3232 bronze badges

$endgroup$

share|cite|improve this answer

answered Mar 28 at 3:43

Thomas ShelbyThomas Shelby

6,39933 gold badges99 silver badges3232 bronze badges

answered Mar 28 at 3:43

Thomas ShelbyThomas Shelby

6,39933 gold badges99 silver badges3232 bronze badges

answered Mar 28 at 3:43

Thomas ShelbyThomas Shelby

6,39933 gold badges99 silver badges3232 bronze badges