Mixing
Integrate $int_{-1}^1 frac x{e^x+x+1}$
$begingroup$
How to approach this integral?I tried to add subtract $e^{x}+1$. But I didn't get too far.
$$int_{-1}^{1}frac{x}{e^{x}+x+1}dx$$
real-analysis integration definite-integrals
edited Jan 12 at 22:12
Michael Rozenberg
103k1891195
asked Jan 12 at 21:38
Vali ROVali RO
736
$endgroup$
add a comment |
$begingroup$
How to approach this integral?I tried to add subtract $e^{x}+1$. But I didn't get too far.
$$int_{-1}^{1}frac{x}{e^{x}+x+1}dx$$
real-analysis integration definite-integrals
edited Jan 12 at 22:12
Michael Rozenberg
103k1891195
asked Jan 12 at 21:38
Vali ROVali RO
736
$endgroup$
add a comment |
$begingroup$
How to approach this integral?I tried to add subtract $e^{x}+1$. But I didn't get too far.
$$int_{-1}^{1}frac{x}{e^{x}+x+1}dx$$
real-analysis integration definite-integrals
edited Jan 12 at 22:12
Michael Rozenberg
103k1891195
asked Jan 12 at 21:38
Vali ROVali RO
736
$endgroup$
How to approach this integral?I tried to add subtract $e^{x}+1$. But I didn't get too far.
$$int_{-1}^{1}frac{x}{e^{x}+x+1}dx$$
real-analysis integration definite-integrals
real-analysis integration definite-integrals
edited Jan 12 at 22:12
Michael Rozenberg
103k1891195
asked Jan 12 at 21:38
Vali ROVali RO
736
edited Jan 12 at 22:12
Michael Rozenberg
103k1891195
asked Jan 12 at 21:38
Vali ROVali RO
736
edited Jan 12 at 22:12
Michael Rozenberg
103k1891195
edited Jan 12 at 22:12
Michael Rozenberg
103k1891195
edited Jan 12 at 22:12
Michael Rozenberg
103k1891195
103k1891195
asked Jan 12 at 21:38
Vali ROVali RO
736
asked Jan 12 at 21:38
Vali ROVali RO
736
asked Jan 12 at 21:38
Vali ROVali RO
736
736
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The hint:
Use $$frac{x}{e^x+x+1}=1-frac{e^x+1}{e^x+x+1}.$$
answered Jan 12 at 21:40
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@Vali RO You are welcome!
$endgroup$
– Michael Rozenberg
Jan 12 at 21:47
add a comment |
$begingroup$
Hint:
$$begin{align}int_{-1}^1frac{x}{e^x+1+x}mathrm dx&=int_{-1}^1frac{x+e^x+1-e^x-1}{e^x+1+x}mathrm dx\&=int_{-1}^1left(1-frac{1+e^x}{1+x+e^x}right)mathrm dxend{align}$$
Can you see how the numerator and denominator relate to each other in the new fractional term? How do you integrate such terms?
answered Jan 12 at 21:43
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
I take u=e^x+x so u'=e^x+1 so the second integral is ln|u+1|
$endgroup$
– Vali RO
Jan 12 at 21:46
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@ValiRO yes exactly :)
$endgroup$
– John Doe
Jan 12 at 23:49
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The hint:
Use $$frac{x}{e^x+x+1}=1-frac{e^x+1}{e^x+x+1}.$$
answered Jan 12 at 21:40
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@Vali RO You are welcome!
$endgroup$
– Michael Rozenberg
Jan 12 at 21:47
add a comment |
$begingroup$
The hint:
Use $$frac{x}{e^x+x+1}=1-frac{e^x+1}{e^x+x+1}.$$
answered Jan 12 at 21:40
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@Vali RO You are welcome!
$endgroup$
– Michael Rozenberg
Jan 12 at 21:47
add a comment |
$begingroup$
The hint:
Use $$frac{x}{e^x+x+1}=1-frac{e^x+1}{e^x+x+1}.$$
answered Jan 12 at 21:40
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
The hint:
Use $$frac{x}{e^x+x+1}=1-frac{e^x+1}{e^x+x+1}.$$
answered Jan 12 at 21:40
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 12 at 21:40
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 12 at 21:40
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 12 at 21:40
Michael RozenbergMichael Rozenberg
103k1891195
103k1891195
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@Vali RO You are welcome!
$endgroup$
– Michael Rozenberg
Jan 12 at 21:47
add a comment |
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@Vali RO You are welcome!
$endgroup$
– Michael Rozenberg
Jan 12 at 21:47
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@Vali RO You are welcome!
$endgroup$
– Michael Rozenberg
Jan 12 at 21:47
$begingroup$
@Vali RO You are welcome!
$endgroup$
– Michael Rozenberg
Jan 12 at 21:47
add a comment |
$begingroup$
Hint:
$$begin{align}int_{-1}^1frac{x}{e^x+1+x}mathrm dx&=int_{-1}^1frac{x+e^x+1-e^x-1}{e^x+1+x}mathrm dx\&=int_{-1}^1left(1-frac{1+e^x}{1+x+e^x}right)mathrm dxend{align}$$
Can you see how the numerator and denominator relate to each other in the new fractional term? How do you integrate such terms?
answered Jan 12 at 21:43
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
I take u=e^x+x so u'=e^x+1 so the second integral is ln|u+1|
$endgroup$
– Vali RO
Jan 12 at 21:46
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@ValiRO yes exactly :)
$endgroup$
– John Doe
Jan 12 at 23:49
add a comment |
$begingroup$
Hint:
$$begin{align}int_{-1}^1frac{x}{e^x+1+x}mathrm dx&=int_{-1}^1frac{x+e^x+1-e^x-1}{e^x+1+x}mathrm dx\&=int_{-1}^1left(1-frac{1+e^x}{1+x+e^x}right)mathrm dxend{align}$$
Can you see how the numerator and denominator relate to each other in the new fractional term? How do you integrate such terms?
answered Jan 12 at 21:43
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
I take u=e^x+x so u'=e^x+1 so the second integral is ln|u+1|
$endgroup$
– Vali RO
Jan 12 at 21:46
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@ValiRO yes exactly :)
$endgroup$
– John Doe
Jan 12 at 23:49
add a comment |
$begingroup$
Hint:
$$begin{align}int_{-1}^1frac{x}{e^x+1+x}mathrm dx&=int_{-1}^1frac{x+e^x+1-e^x-1}{e^x+1+x}mathrm dx\&=int_{-1}^1left(1-frac{1+e^x}{1+x+e^x}right)mathrm dxend{align}$$
Can you see how the numerator and denominator relate to each other in the new fractional term? How do you integrate such terms?
answered Jan 12 at 21:43
John DoeJohn Doe
11.1k11238
$endgroup$
Hint:
$$begin{align}int_{-1}^1frac{x}{e^x+1+x}mathrm dx&=int_{-1}^1frac{x+e^x+1-e^x-1}{e^x+1+x}mathrm dx\&=int_{-1}^1left(1-frac{1+e^x}{1+x+e^x}right)mathrm dxend{align}$$
Can you see how the numerator and denominator relate to each other in the new fractional term? How do you integrate such terms?
answered Jan 12 at 21:43
John DoeJohn Doe
11.1k11238
answered Jan 12 at 21:43
John DoeJohn Doe
11.1k11238
answered Jan 12 at 21:43
John DoeJohn Doe
11.1k11238
answered Jan 12 at 21:43
John DoeJohn Doe
11.1k11238
11.1k11238
$begingroup$
I take u=e^x+x so u'=e^x+1 so the second integral is ln|u+1|
$endgroup$
– Vali RO
Jan 12 at 21:46
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@ValiRO yes exactly :)
$endgroup$
– John Doe
Jan 12 at 23:49
add a comment |
$begingroup$
I take u=e^x+x so u'=e^x+1 so the second integral is ln|u+1|
$endgroup$
– Vali RO
Jan 12 at 21:46
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@ValiRO yes exactly :)
$endgroup$
– John Doe
Jan 12 at 23:49
$begingroup$
I take u=e^x+x so u'=e^x+1 so the second integral is ln|u+1|
$endgroup$
– Vali RO
Jan 12 at 21:46
$begingroup$
I take u=e^x+x so u'=e^x+1 so the second integral is ln|u+1|
$endgroup$
– Vali RO
Jan 12 at 21:46
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
Thank you, I did it :)
$endgroup$
– Vali RO
Jan 12 at 21:47
$begingroup$
@ValiRO yes exactly :)
$endgroup$
– John Doe
Jan 12 at 23:49
$begingroup$
@ValiRO yes exactly :)
$endgroup$
– John Doe
Jan 12 at 23:49
add a comment |
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