Presumably what madhukar means is that this integral probably does not have a closed form solution. I agree - I don't see a good way to integrate it by parts.
You can find a power series for it by replace x=e^u and writing the intregral of:
e^(2u)/u du.
Then e^(2u) = sum (n=0,...,oo) (2u)^n/n!
= 1 + sum(n=1,...,oo) (2u)^n/n!
Dividing by u:
e^(2u)/u = 1/u + sum(n=1,oo) 2^n . u^(n-1)/n!
Integrating each term we get:
integral of e^(2u)/u = ln u + sum(n=1,oo) (2u)^n/(n.n!)
Putting u=ln x back into the equation, you get:
integral = ln ln x + sum (n=1,oo) (2 log x)^n/(n.n!)
2007-10-30 07:42:03
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answer #1
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answered by thomasoa 5
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i'm uncertain no count if that's ( a million/x)*ln x³ ... or ... a million / ( x. ln x³ ). consequently, I honestly have presented the two possible circumstances : ......................................... [ a million ] ? ( a million/x ). ln ( x³ ) dx = 3* ? ( ln x ) * ( a million/x ) dx ......(a million) ......................................... positioned u = ln x ... so as that ... du = ( a million/x ) dx. ......................................... From (a million), then, I = 3 * ? u du = 3 * ( u² / 2 ) + C = ( 3/2 )* ( ln x )² + C. ....................Ans. ......................................... [ 2 ] I = ? [ a million / ( x. ln ( x³ ) ) ] d = ? [ a million / ( 3* x. ln x ) ] dx = ( a million/3 ) * ? [ a million / ( ln x ) ] * ( a million/x ) dx = ( a million/3 ) * ? ( a million/u ) du, .......................the place u = ln x = ( a million/3 ) * ln u + C = ( a million/3 )* ln ( ln x ) + C. ............Ans. ......................................... chuffed to help ! .........................................
2016-12-30 10:45:50
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answer #2
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answered by bockoven 3
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I doubt if it is integrable. Every function is not integrable.
2007-10-30 07:24:05
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answer #3
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answered by Madhukar 7
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(1.lnx - x. 1/x) / (ln x)^2
= ( ln x - 1) / ( ln x)^2
2007-10-30 07:18:05
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answer #4
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answered by pride_rock21 2
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what are the bounds?
2007-10-30 07:10:58
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answer #5
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answered by ¥€$ 3
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