How to solve $intfrac{ln x}{x^2(ln (x)-1)^2}dx$ by substitution
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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 indefinite-integrals
edited 18 hours ago
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up vote
5
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 indefinite-integrals
edited 18 hours ago
user21820
38k541149
asked yesterday
Andres Oropeza
305
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
5
down vote
favorite
up vote
5
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 indefinite-integrals
edited 18 hours ago
user21820
38k541149
asked yesterday
Andres Oropeza
305
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 indefinite-integrals
calculus integration indefinite-integrals
edited 18 hours ago
user21820
38k541149
asked yesterday
Andres Oropeza
305
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 18 hours ago
user21820
38k541149
asked yesterday
Andres Oropeza
305
New contributor
Andres Oropeza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 18 hours ago
user21820
38k541149
edited 18 hours ago
user21820
38k541149
edited 18 hours ago
user21820
38k541149
38k541149
asked yesterday
Andres Oropeza
305
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 yesterday
Andres Oropeza
305
asked yesterday
Andres Oropeza
305
305
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.
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1 Answer
1
active
oldest
votes
up vote
15
down vote
accepted
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered yesterday
Yadati Kiran
956316
5
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
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered yesterday
Yadati Kiran
956316
5
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
yesterday
add a comment |
up vote
15
down vote
accepted
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered yesterday
Yadati Kiran
956316
5
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
yesterday
add a comment |
up vote
15
down vote
accepted
up vote
15
down vote
accepted
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered yesterday
Yadati Kiran
956316
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered yesterday
Yadati Kiran
956316
answered yesterday
Yadati Kiran
956316
answered yesterday
Yadati Kiran
956316
answered yesterday
Yadati Kiran
956316
956316
5
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
yesterday
add a comment |
5
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
yesterday
5
5
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
yesterday
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
yesterday
add a comment |
Andres Oropeza is a new contributor. Be nice, and check out our Code of Conduct.
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