Find:
\int \left ( \log x \right )^{2}dx

 

 

 

 
 
 
 
 

Answers (1)

I= \int \left ( \log x \right )^{2}dx \: \: \: \: put\: \log x= t
  I= \int t^{2}\cdot e^{t}dt\: \: \: \: \: \: \: \: \frac{1}{x}dx= dt
                                    \Rightarrow dx= xdt
                                    \Rightarrow dx= e^{t}dt\: \: \left [ \because x= e^{t} \right ]
\Rightarrow t^{2}\int e^{t}dt-\int \left ( \frac{d}{dt}\left ( t^{2} \right )\int e^{t}dt \right )dt
\Rightarrow t^{2}\: e^{t}-\int \left ( 2t\, e^{t} \right )dt\:\: \Rightarrow \: t^{2}e^{t}-2\int te^{t}dt
\Rightarrow t^{2}e^{t}-2\int t\: e^{t}dt\: \: \Rightarrow\: \: t^{2}e^{t}-2\left [ te^{t} -\int 1\cdot e^{t}dt\right ]
\Rightarrow t^{2}e^{t}-2\left [ t\, e^{t}-e^{t} \right ]+C\: \: \Rightarrow \: \: t^{2}e^{t}-2te^{t}+2e^{t}+C
        substituting  t= \log x
\Rightarrow \left ( \log x \right )^{2}e^{\log x}-2\log x\, \: e^{\log x}+2\: e^{\log x}+C
\Rightarrow \left ( \log x \right )^{2}\left ( x \right )-2\log x\: \left ( x \right )+2x+C
\Rightarrow \left x( \log x \right )^{2}-2x\log x+2x+ C  

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