# What is the integration of x log x dx ?

## Solution :

We have, I = $$\int$$ x log x dx

By using integration by parts,

And taking log x as first function and x as second function. Then,

I = log x { $$\int$$ x dx } – $$\int$$ { $${d\over dx}(log x) \times \int x dx$$ } dx

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

$$\implies$$ I = $$x^2\over 2$$ log x – $$1\over 2$$ $$\int$$ x dx

$$\implies$$ I = $$x^2\over 2$$ log x – $$1\over 2$$ ($$x^2\over 2$$) + C

$$\implies$$ I = $$x^2\over 2$$ log x – $$x^2\over 2$$ + C

Hence. the integration of x log x with respect to x is $$x^2\over 2$$ log x – $$x^2\over 2$$ + C

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