Indefinite Integration

Let $\large t=\ln { x }$ ,

Also $\large dt=\frac { dx }{ x }$ .

Rearranging the above expression we get, $\large dx={ e }^{ t }dt$ .

$\large \int { (\frac { 1 }{ \ln { x } } -\frac { 1 }{ { \ln { x } }^{ 2 } } )dx }$ .

Substituting, $\large \int { { e }^{ t }(\frac { 1 }{ t } -\frac { 1 }{ { t }^{ 2 } } ) } dt$

Using the property, $\int { { e }^{ x }(f\left( x \right) +f^{ ' }\left( x \right) ) } dx={ e }^{ x }f\left( x \right)$ . (Proof of the above property is trivial and is left as an exercise to the reader.)

Therefore we get, $\large \frac { { e }^{ t } }{ t }$

Bringing $x$ back, $\large \frac { { e }^{ \ln { x } } }{ \ln { x } } +c$

Simplifying we get, $\large \boxed{\frac { x }{ \ln { x } } +c}$ .

By inspection we find that:

$\dfrac {d}{dx} \left( \dfrac {x}{\ln {x}} \right) = \dfrac {1}{\ln{x}} + \dfrac {x(-1)}{x(\ln {x})^2} = \dfrac {1}{\ln{x}} - \dfrac {1}{(\ln {x})^2}$

$\displaystyle \Rightarrow \int {\left(\dfrac {1}{\ln{x}} - \dfrac {1}{(\ln {x})^2}\right)} = \boxed{\dfrac {x}{\ln {x}} + c}$

If we use integration by parts on first part, second part will cancel out and we get the required result.

In manual integration:

$\begin{aligned} \int \left(\frac{1}{\ln(x)} - \frac{1}{(\ln(x))^2} \right)\,dx &= -\int \frac{1}{\ln^2(x)} - \int -\frac{1}{\ln^2(x)} +\frac{x}{\ln(x)} \quad\quad\quad\quad\quad\quad\quad\quad{\text{Use integration by parts}} \\& = \frac{x}{\ln(x)} + C \end{aligned}$