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```
Integrate:- {(1/(logx))-(1/(logx) 2 )} for upper limit e 2 and lower limit 0.

```
9 years ago

```							I = {1/logx -1/(logx)2 }dx
multiply numerator & denominator by x
I = x{1/logx-1/(logx)2}dx /x
put logx = t
dx/x = dt      & x=et
now
I = et{1/t - 1/t2}dt
I = et{ f(t) + f1(t) }
we have a direct formula for ex{f(x) + f1(x)}dx = exf(x)
so I = et f(t)
=et/t = (x / logx)       lim 0 to e2
=0 - e2/2
```
9 years ago
```							Take logx= t and x=e^tdx=e^t × dt So integral becomes, integration [1÷t- 1÷t^2]× e^tThis is of the form e^x( f(x) +f'(x) )= e^x(f(x) )+c So answer becomes e^logx[ 1÷logx] +c
```
2 years ago
```							Integration dx/log x - integration dx/(log x)^2Intigrating the first part by parts formula1/log x integration 1 dx - integration ( d/dx (log x)^-1.integration 1 dx ) dx - integration dx/(log x)^2x/log x  + integration dx/(log x)^2 - integration dx/(log x)^2= x/log x ans.
```
2 years ago 605 Points
```							Dear student,Please find the solution to your problem below. I = {1/logx -1/(logx)2 }dxmultiply numerator & denominator by xI = x{1/logx-1/(logx)2}dx /xput logx = t    dx/x = dt      & x=etnow        I = et{1/t - 1/t2}dt               I = et{ f(t) + f1(t) }we have a direct formula for ex{f(x) + f1(x)}dx = exf(x)so I = et f(t)         = et/t = (x / logx)       lim 0 to e2        = 0 – e2/2        = – e2/2 Thanks and regards,Kushagra
```
4 months ago
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