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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?
$begingroup$
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 find this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $
calculus integration indefinite-integrals
$endgroup$
|
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 find this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $
calculus integration indefinite-integrals
$endgroup$
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 $$f(x) = int e^x biggr(arctan x + frac 2x(1+x^2)^2biggr),dx$$ 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 find this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $
calculus integration indefinite-integrals
$endgroup$
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 find this result quickly? It seems very complicated. The answer is $e(pi/4-(1/2)). $
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited 1 hour ago
Hema
asked 4 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
|
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
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
|
show 6 more comments
2 Answers
2
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 $fracepi4 - frace2 +1$.
$endgroup$
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 $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
oldest
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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 $fracepi4 - frace2 +1$.
$endgroup$
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 $fracepi4 - frace2 +1$.
$endgroup$
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 $fracepi4 - frace2 +1$.
$endgroup$
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 3 hours ago
Catalin ZaraCatalin Zara
3,817514
3,817514
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 $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$
$endgroup$
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 $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$
$endgroup$
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 $$f (1)=biggr[e^x biggr(arctan x - frac 11+x^2biggr)biggr]_0^1=frac epi4-frac e2+1. $$
$endgroup$
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 2 hours ago
Thomas ShelbyThomas Shelby
4,4992726
4,4992726
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
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