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!!!