Question

Evaluate the following integral with complete steps

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

Answer 1

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

Given integral is

$\rm \: \displaystyle\int\rm \bigg(log(logx) + \dfrac{1}{ {(logx)}^{2} } \bigg) \: dx$

To solve this integral, we use method of substitution.

So, substituting

$\rm \: logx = y$

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

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

So, on substituting all these values, we get

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

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

can be further rewritten as

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

We know,

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

So, for the first integral,

$\rm \: f(y) = logy$

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

And, for the second integral,

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

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

So, using the above formula, we get

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

$\rm \: = \: {e}^{y}\bigg(logy - \dfrac{1}{y} \bigg) + c$

$\rm \: = \: x\bigg(log(logx) - \dfrac{1}{logx} \bigg) + c$

Hence,

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

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

Remark :- Proof of the property

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

Consider,

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

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

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

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

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

$\rm \: = \: {e}^{x}f(x)\: + \: c$

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

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

Step-by-step explanation:

$\text{Let \( \displaystyle\rm \quad I=\int\left\{\log (\log x)+\frac{1}{(\log x)^{2}}\right\} d x \)}$

$\text{Let Put } \rm \log x=t$

$\[ \begin{array}{rlr} \displaystyle\rm \Rightarrow x & \displaystyle\rm=e^{t} \quad \Rightarrow d x=e^{t} d t \\ \\ & \displaystyle\rm=\int\left(\log t+\frac{1}{t^{2}}\right) e^{t} d t \\ \\ & \displaystyle\rm=\int e^{t} \cdot\left(\log t+\frac{1}{t}-\frac{1}{t}+\frac{1}{t^{2}}\right) d t \quad \quad \text { (Note this step) }\\ \\ &\displaystyle\rm =\int e^{t} \cdot\left(\log t+\frac{1}{t}\right) d t-\int e^{t}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) d t \\ \\ & \displaystyle\rm=e^{t} \cdot \log t-e^{t} \cdot \frac{1}{t}+c \\ \\ & \displaystyle\rm=x \cdot \log (\log x)-x \cdot \frac{1}{\log x}+c\\ \\ & \boxed{\color{violet} \displaystyle\rm=x\left\{\log (\log x)-\frac{1}{\log x}\right\}+c }\quad \quad\left(\begin{array}{c}\displaystyle\rm \because t=\log x \\\displaystyle\rm x=e^{t} \end{array}\right) \end{array} \]$