How to solve this integral by u-substitution
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3
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$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$
Hello, I haven't be able to solve this integral, I've tried to do $u = ln (x)-1$ but couldn't make it work, any insight?
calculus integration
edited 1 hour ago
KM101
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asked 2 hours ago
Andres Oropeza
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up vote
3
down vote
favorite
$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$
Hello, I haven't be able to solve this integral, I've tried to do $u = ln (x)-1$ but couldn't make it work, any insight?
calculus integration
edited 1 hour ago
KM101
2,061314
asked 2 hours ago
Andres Oropeza
161
New contributor
Andres Oropeza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$
Hello, I haven't be able to solve this integral, I've tried to do $u = ln (x)-1$ but couldn't make it work, any insight?
calculus integration
edited 1 hour ago
KM101
2,061314
asked 2 hours ago
Andres Oropeza
161
New contributor
Andres Oropeza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$
Hello, I haven't be able to solve this integral, I've tried to do $u = ln (x)-1$ but couldn't make it work, any insight?
calculus integration
calculus integration
edited 1 hour ago
KM101
2,061314
asked 2 hours ago
Andres Oropeza
161
New contributor
Andres Oropeza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
KM101
2,061314
asked 2 hours ago
Andres Oropeza
161
New contributor
Andres Oropeza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
KM101
2,061314
edited 1 hour ago
KM101
2,061314
edited 1 hour ago
KM101
2,061314
2,061314
asked 2 hours ago
Andres Oropeza
161
New contributor
Andres Oropeza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Andres Oropeza
161
asked 2 hours ago
Andres Oropeza
161
161
New contributor
Andres Oropeza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Andres Oropeza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Andres Oropeza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
7
down vote
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered 1 hour ago
Yadati Kiran
728214
2
I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
– YiFan
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered 1 hour ago
Yadati Kiran
728214
2
I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
– YiFan
1 hour ago
add a comment |
up vote
7
down vote
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered 1 hour ago
Yadati Kiran
728214
2
I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
– YiFan
1 hour ago
add a comment |
up vote
7
down vote
up vote
7
down vote
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered 1 hour ago
Yadati Kiran
728214
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered 1 hour ago
Yadati Kiran
728214
answered 1 hour ago
Yadati Kiran
728214
answered 1 hour ago
Yadati Kiran
728214
answered 1 hour ago
Yadati Kiran
728214
728214
2
I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
– YiFan
1 hour ago
add a comment |
2
I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
– YiFan
1 hour ago
2
2
I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
– YiFan
1 hour ago
I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
– YiFan
1 hour ago
add a comment |
Andres Oropeza is a new contributor. Be nice, and check out our Code of Conduct.
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