Question

Integrate the following

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

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm \frac{logx - 1}{ {(logx)}^{2} } \: dx$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\rm \: logx \: = \: y$

$\rm\implies \:x = {e}^{y}$

$\rm\implies \:dx = {e}^{y} \: dy$

So, on substituting these values in given integral, we get

$\rm \: =  \:\displaystyle\int\rm \frac{y - 1}{ {y}^{2} } \: {e}^{y} \: dy$

can be further rewritten as

$\rm \: =  \:\displaystyle\int\rm \bigg(\dfrac{y}{ {y}^{2} } - \dfrac{1}{ {y}^{2} } \bigg) {e}^{y} \: dy$

$\rm \: =  \:\displaystyle\int\rm \bigg(\dfrac{1}{ {y}} - \dfrac{1}{ {y}^{2} } \bigg) {e}^{y} \: dy$

We know,

$\boxed{\rm{  \:\displaystyle\int\rm {e}^{x}[f(x) + f'(x)] \: dx \: = \: {e}^{x}f(x) + c \: \: }}$

So, here in this integral,

$\rm \: f(y) = \dfrac{1}{y}$

$\rm \: f'(y) = - \: \dfrac{1}{ {y}^{2} }$

So, using above result, we get

$\rm \: =  \:{e}^{y} \: \times \dfrac{1}{y} + c$

$\rm \: =  \:\dfrac{x}{logx} + c$

Hence,

$\boxed{\rm{  \:\rm \: \displaystyle\int\rm \frac{logx - 1}{ {(logx)}^{2} } \: dx \: = \: \frac{x}{logx} + c \: \:}}$

$\rule{190pt}{2pt}$

Proof of

$\boxed{\rm{  \:\displaystyle\int\rm {e}^{x}[f(x) + f'(x)] \: dx \: = \: {e}^{x}f(x) + c \: \: }}$

Consider,

$\rm\:\displaystyle\int\rm {e}^{x}[f(x) + f'(x)] \: dx \:$

can be rewritten as

$\rm \: =  \:\displaystyle\int\rm {e}^{x} \: f(x)dx \: + \: \displaystyle\int\rm {e}^{x}f'(x)dx$

Now, using integration by parts in first integral, we get

$\rm \: =  \:f(x)\displaystyle\int\rm {e}^{x} \: dx - \displaystyle\int\rm \bigg[\dfrac{d}{dx}f(x)\displaystyle\int\rm {e}^{x}dx\bigg]dx+ \: \displaystyle\int\rm {e}^{x}f'(x)dx$

$\rm \: =  \:{e}^{x}f(x) - \displaystyle\int\rm f'(x){e}^{x} \: dx+ \: \displaystyle\int\rm {e}^{x}f'(x)dx + c$

$\rm \: =  \:{e}^{x}f(x) + c$

Hence,

$\rm\implies \:\rm\:\displaystyle\int\rm {e}^{x}[f(x) + f'(x)] \: dx \: = \: {e}^{x}f(x) \: + \: c$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

SOLUTION :

please check the attached file