Shortcut for value of $f(1)$ where $f(x) = int e^x left(arctan x + frac 2x(1+x^2)^2right),dx$ The Next CEO of Stack Overflow$int sqrtfracxx+1dx$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$$int frac2x9x^2+3dx=?$Integrate $int frac11+arctan(x)dx$A shortcut for solving $int fracdx(sin x+cos x)^3$Calculate $int_0^1fraccos(arctan x)sqrtxdx$About the integral $intarctanleft(frac1sinh^2 xright)dx$, some idea or feedbackFind $intarctan x,mathrm dx$ without substitutionHow can I solve $int_0^1fracarctan(x^2)1+x^2,mathrm dx$?
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Shortcut for value of $f(1)$ where $f(x) = int e^x left(arctan x + frac 2x(1+x^2)^2right),dx$
The Next CEO of Stack Overflow$int sqrtfracxx+1dx$Integrate $ int frac e^arctan(x)(1+x^2)^frac32 dx $Problem with definite integral $int _0^fracpi 2sin left(arctan left(xright)+xright)dx$$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 substitutionHow can I solve $int_0^1fracarctan(x^2)1+x^2,mathrm dx$?
$begingroup$
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
edited Mar 28 at 8:09
YuiTo Cheng
2,1862937
asked Mar 28 at 2:28
HemaHema
6621213
$endgroup$
|
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
edited Mar 28 at 8:09
YuiTo Cheng
2,1862937
asked Mar 28 at 2:28
HemaHema
6621213
$endgroup$
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
edited Mar 28 at 8:09
YuiTo Cheng
2,1862937
asked Mar 28 at 2:28
HemaHema
6621213
$endgroup$
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 finding this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Mar 28 at 8:09
YuiTo Cheng
2,1862937
asked Mar 28 at 2:28
HemaHema
6621213
edited Mar 28 at 8:09
YuiTo Cheng
2,1862937
asked Mar 28 at 2:28
HemaHema
6621213
edited Mar 28 at 8:09
YuiTo Cheng
2,1862937
edited Mar 28 at 8:09
YuiTo Cheng
2,1862937
edited Mar 28 at 8:09
YuiTo Cheng
2,1862937
2,1862937
asked Mar 28 at 2:28
HemaHema
6621213
asked Mar 28 at 2:28
HemaHema
6621213
asked Mar 28 at 2:28
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
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
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
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
Mar 28 at 2:50
$begingroup$
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. 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
$begingroup$
@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
$begingroup$
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
$begingroup$
@amsmath Joint Entrance Exam in India. en.wikipedia.org/wiki/Joint_Entrance_Examination
$endgroup$
– Deepak
Mar 28 at 3:47
|
show 6 more comments
2 Answers
2
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$begingroup$
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 $fracepi4 - frace2 +1$.
answered Mar 28 at 2:51
Catalin ZaraCatalin Zara
3,827514
$endgroup$
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 $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$
answered Mar 28 at 3:43
Thomas ShelbyThomas Shelby
4,5862727
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
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 $fracepi4 - frace2 +1$.
answered Mar 28 at 2:51
Catalin ZaraCatalin Zara
3,827514
$endgroup$
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 $fracepi4 - frace2 +1$.
answered Mar 28 at 2:51
Catalin ZaraCatalin Zara
3,827514
$endgroup$
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 $fracepi4 - frace2 +1$.
answered Mar 28 at 2:51
Catalin ZaraCatalin Zara
3,827514
$endgroup$
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 $fracepi4 - frace2 +1$.
answered Mar 28 at 2:51
Catalin ZaraCatalin Zara
3,827514
answered Mar 28 at 2:51
Catalin ZaraCatalin Zara
3,827514
answered Mar 28 at 2:51
Catalin ZaraCatalin Zara
3,827514
answered Mar 28 at 2:51
Catalin ZaraCatalin Zara
3,827514
3,827514
add a comment |
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 $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$
answered Mar 28 at 3:43
Thomas ShelbyThomas Shelby
4,5862727
$endgroup$
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 $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$
answered Mar 28 at 3:43
Thomas ShelbyThomas Shelby
4,5862727
$endgroup$
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 $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$
answered Mar 28 at 3:43
Thomas ShelbyThomas Shelby
4,5862727
$endgroup$
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 $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$
answered Mar 28 at 3:43
Thomas ShelbyThomas Shelby
4,5862727
answered Mar 28 at 3:43
Thomas ShelbyThomas Shelby
4,5862727
answered Mar 28 at 3:43
Thomas ShelbyThomas Shelby
4,5862727
answered Mar 28 at 3:43
Thomas ShelbyThomas Shelby
4,5862727
4,5862727
add a comment |
add a comment |
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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
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