# Integrate:-{(1/(logx))-(1/(logx)2)} for upper limit e2 and lower limit 0.

509 Points
12 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

Manav
13 Points
5 years ago
Take logx= t and x=e^t
dx=e^t × dt
So integral becomes, integration [1÷t- 1÷t^2]× e^t
This is of the form e^x( f(x) +f'(x) )= e^x(f(x) )+c
So answer becomes e^logx[ 1÷logx] +c
15 Points
4 years ago
Integration dx/log x - integration dx/(log x)^2
Intigrating the first part by parts formula
1/log x integration 1 dx - integration ( d/dx (log x)^-1.integration 1 dx ) dx - integration dx/(log x)^2
x/log x  + integration dx/(log x)^2 - integration dx/(log x)^2= x/log x ans.
3 years ago
Dear student,

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
= – e2/2

Thanks and regards,
Kushagra