Question

Evaluate:
$\int (\frac{1}{\log x}-\frac{1}{{\left(\log x\right)}^2})dx$

Answer 1

$\int \frac{1}{logx}-\frac{1}{(logx{)}^{{}^2}}$

$\Rightarrow Letlogx=t$

$x=e^t.$ $[dx=e^{t}dt]$

then

$\Rightarrow \int e^t[\frac{1}{t}-\frac{1}{t^2}]dt----\left(1\right)$

we know $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]=e^x=f)x$

and

we know $\frac{d}{dt}$ $\left(\frac{1}{t}\right)$=$\frac{-1}{t^2}$

Form

eq(1) we have

$\Rightarrow \int e^t[\frac{1}{t}-\frac{1}{t^2}]dt$ = $e^t\left(\frac{1}{t}\right)$

$\Rightarrow e^{logx}.\frac{1}{logx}$