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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

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

edited 1 hour ago

Hema

asked 3 hours ago

HemaHema

6531213

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

 | 
show 6 more comments

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

edited 1 hour ago

Hema

asked 3 hours ago

HemaHema

6531213

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

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

edited 1 hour ago

Hema

asked 3 hours ago

HemaHema

6531213

$endgroup$

share|cite|improve this question

edited 1 hour ago

Hema

asked 3 hours ago

HemaHema

6531213

share|cite|improve this question

edited 1 hour ago

Hema

asked 3 hours ago

HemaHema

6531213

asked 3 hours ago

HemaHema

6531213

asked 3 hours ago

HemaHema

6531213

2 Answers
2

active

oldest

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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

answered 3 hours ago

Catalin ZaraCatalin Zara

3,817514

$endgroup$

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

answered 2 hours ago

Thomas ShelbyThomas Shelby

4,4992726

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add a comment | 

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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

answered 3 hours ago

Catalin ZaraCatalin Zara

3,817514

$endgroup$

add a comment | 

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

answered 3 hours ago

Catalin ZaraCatalin Zara

3,817514

$endgroup$

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

share|cite|improve this answer

answered 3 hours ago

Catalin ZaraCatalin Zara

3,817514

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Catalin ZaraCatalin Zara

3,817514

answered 3 hours ago

Catalin ZaraCatalin Zara

3,817514

answered 3 hours ago

Catalin ZaraCatalin Zara

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

answered 2 hours ago

Thomas ShelbyThomas Shelby

4,4992726

$endgroup$

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

answered 2 hours ago

Thomas ShelbyThomas Shelby

4,4992726

$endgroup$

add a comment | 

$begingroup$

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

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

answered 2 hours ago

Thomas ShelbyThomas Shelby

4,4992726

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Thomas ShelbyThomas Shelby

4,4992726

answered 2 hours ago

Thomas ShelbyThomas Shelby

4,4992726

answered 2 hours ago

Thomas ShelbyThomas Shelby

4,4992726