# 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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