integral [ 1/log x - 1/ (log x)2 ]dx??

 we need to find 

∫1/logx  -   1/(logx)2   dx ..


now, take logx = t

=> x = et

=> dx = et dt

=> the given integral becomes,

∫( 1/t  -  1/t2  ) et dt

=  ∫ et (1/t  -  1/t2)  dt


now, we have a formula :

∫ex ( f(x) + f'(x) ) dx  =  ex f(x) ............ ............  (1)


so here, let f(x)  = 1/t  => f'(x) = derivative of f(x)  =  -1/t2.

so, it is of the form.. (1)


so the given integral = ex f(x) = et (1/t) + C

substituting the value of t we get, 

answer = elogx (1/logx)  +  C  =  x / logx + C


hope it helps!!

cheers!!!

  • 92
What are you looking for?