How to solve $intfrac{ln x}{x^2(ln (x)-1)^2}dx$ by substitution
7
$begingroup$
$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$
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 '18 at 6:15
user21820
38.9k543153
asked Nov 23 '18 at 21:05
Andres OropezaAndres Oropeza
405
$endgroup$
add a comment |
7
$begingroup$
$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$
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 '18 at 6:15
user21820
38.9k543153
asked Nov 23 '18 at 21:05
Andres OropezaAndres Oropeza
405
$endgroup$
add a comment |
7
7
7
1
$begingroup$
$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$
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 '18 at 6:15
user21820
38.9k543153
asked Nov 23 '18 at 21:05
Andres OropezaAndres Oropeza
405
$endgroup$
$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$
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 '18 at 6:15
user21820
38.9k543153
asked Nov 23 '18 at 21:05
Andres OropezaAndres Oropeza
405
edited Nov 24 '18 at 6:15
user21820
38.9k543153
asked Nov 23 '18 at 21:05
Andres OropezaAndres Oropeza
405
edited Nov 24 '18 at 6:15
user21820
38.9k543153
edited Nov 24 '18 at 6:15
user21820
38.9k543153
edited Nov 24 '18 at 6:15
user21820
38.9k543153
38.9k543153
asked Nov 23 '18 at 21:05
Andres OropezaAndres Oropeza
405
asked Nov 23 '18 at 21:05
Andres OropezaAndres Oropeza
405
asked Nov 23 '18 at 21:05
Andres OropezaAndres Oropeza
405
405
add a comment |
add a comment |
1 Answer
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16
$begingroup$
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered Nov 23 '18 at 21:20
Yadati KiranYadati Kiran
1,7851619
$endgroup$
5
$begingroup$
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.
$endgroup$
– YiFan
Nov 23 '18 at 21:39
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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16
$begingroup$
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered Nov 23 '18 at 21:20
Yadati KiranYadati Kiran
1,7851619
$endgroup$
5
$begingroup$
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.
$endgroup$
– YiFan
Nov 23 '18 at 21:39
add a comment |
16
$begingroup$
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered Nov 23 '18 at 21:20
Yadati KiranYadati Kiran
1,7851619
$endgroup$
5
$begingroup$
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.
$endgroup$
– YiFan
Nov 23 '18 at 21:39
add a comment |
16
16
16
$begingroup$
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered Nov 23 '18 at 21:20
Yadati KiranYadati Kiran
1,7851619
$endgroup$
$xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$
answered Nov 23 '18 at 21:20
Yadati KiranYadati Kiran
1,7851619
answered Nov 23 '18 at 21:20
Yadati KiranYadati Kiran
1,7851619
answered Nov 23 '18 at 21:20
Yadati KiranYadati Kiran
1,7851619
answered Nov 23 '18 at 21:20
Yadati KiranYadati Kiran
1,7851619
1,7851619
5
$begingroup$
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.
$endgroup$
– YiFan
Nov 23 '18 at 21:39
add a comment |
5
$begingroup$
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.
$endgroup$
– YiFan
Nov 23 '18 at 21:39
5
5
$begingroup$
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.
$endgroup$
– YiFan
Nov 23 '18 at 21:39
$begingroup$
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.
$endgroup$
– YiFan
Nov 23 '18 at 21:39
add a comment |
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