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 Nov 24 at 6:15
user21820
38.1k541150
asked Nov 23 at 21:05
Andres Oropeza
305
add a comment |
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 Nov 24 at 6:15
user21820
38.1k541150
asked Nov 23 at 21:05
Andres Oropeza
305
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 Nov 24 at 6:15
user21820
38.1k541150
asked Nov 23 at 21:05
Andres Oropeza
305
$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$
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 Nov 24 at 6:15
user21820
38.1k541150
asked Nov 23 at 21:05
Andres Oropeza
305
edited Nov 24 at 6:15
user21820
38.1k541150
asked Nov 23 at 21:05
Andres Oropeza
305
edited Nov 24 at 6:15
user21820
38.1k541150
edited Nov 24 at 6:15
user21820
38.1k541150
edited Nov 24 at 6:15
user21820
38.1k541150
38.1k541150
asked Nov 23 at 21:05
Andres Oropeza
305
asked Nov 23 at 21:05
Andres Oropeza
305
asked Nov 23 at 21:05
Andres Oropeza
305
305
add a comment |
add a comment |
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 Nov 23 at 21:20
Yadati Kiran
1,282317
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
Nov 23 at 21:39
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 Nov 23 at 21:20
Yadati Kiran
1,282317
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
Nov 23 at 21:39
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 Nov 23 at 21:20
Yadati Kiran
1,282317
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
Nov 23 at 21:39
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 Nov 23 at 21:20
Yadati Kiran
1,282317
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered Nov 23 at 21:20
Yadati Kiran
1,282317
answered Nov 23 at 21:20
Yadati Kiran
1,282317
answered Nov 23 at 21:20
Yadati Kiran
1,282317
answered Nov 23 at 21:20
Yadati Kiran
1,282317
1,282317
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
Nov 23 at 21:39
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
Nov 23 at 21:39
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
Nov 23 at 21:39
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
Nov 23 at 21:39
add a comment |
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