Math, asked by radhika2468, 3 days ago

Solve the following question of integration

Integrate (logx - 1)/(logx)^2

Answers

Answered by mathdude500
6

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

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

So, on substituting these values, we get

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

can be rewritten as

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

\rm \:  =  \: \displaystyle\int\rm  \bigg( \frac{1}{y} -  \frac{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 \: }} \\

Here,

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

and

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

So, using above result, we get

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

\rm \:  =  \:  x  \times \dfrac{1}{logx}  + c \\

\rm \:  =  \: \dfrac{x}{logx}  + c \\

Hence,

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

\rule{190pt}{2pt}

Proof of the result :-

\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, on applying by parts in first integral, we have

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

Answered by talpadadilip417
1

Step-by-step explanation:

 \\  \text{Let \( \displaystyle \tt I=\int \dfrac{\log x-1}{(\log x)^{2}} d x \)}

 \text{Put \( \tt \log x=t  \:  \:  \:  \:  \:  \:  \: \Rightarrow x=e^{t} \quad \therefore d x=e^{t} d t \)}

\[ \begin{aligned} \tt \therefore \quad I & \tt=\int \frac{t-1}{t^{2}} \cdot e^{t} d t=\int e^{t}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) d t \\  \\ & \tt=\int \frac{1}{t} \cdot e^{t} d t-\int \frac{1}{t} \cdot e^{t} d t \end{aligned} \] \\

Integrating first integral by parts, we get

 \\ \[ \begin{aligned} \tt I & \tt=\frac{1}{t} \cdot e^{t}-\int-\frac{1}{t^{2}} \cdot e^{t} d t-\int \frac{1}{t^{2}} \cdot e^{t} d t \\ \\  & \tt=\frac{1}{t} \cdot e^{t}+C \\   \\ &   \color{orangered}\tt=\frac{x}{\log x}+C \end{aligned} \]

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