What is the integration of \((log x)^2\) dx ?

Solution :

We have, I = \((log x)^2\) . 1 dx, Then ,

where \((log x)^2\) is the first function and 1 is the second function according to ilate rule,

I = \((log x)^2\) { \(\int\) 1 dx} – \(\int\) {\(d\over dx\) \((log x)^2\) . \(\int\) 1 dx } dx

I = \((log x)^2\) x – \(\int\) 2 log x . \(1\over x\) . x dx

I = x \((log x)^2\) – 2 \(\int\) log x .1 dx

\(\implies\) I = x \((log x)^2\) – 2[ log x { \(\int\) 1 dx } – \(\int\) { \(d\over dx\) (log x) \(\int\) 1 dx } dx ]

\(\implies\) I = x \((log x)^2\) – 2 { (log x) x – \(\int\) \(1\over x\) x dx }

Hence, I = x( \((log x)^2\) – 2 (x log x – x) + C


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