What is the integration of x tan inverse x dx ?

Solution :

Let I = $$\int$$ x$$tan^{-1}x$$ dx

By using Integration by parts rule,

Taking tan inverse x as first function and x as second function. Then,

I = ($$tan^{-1}x$$) $$\int$$ x dx – $$\int$${$${d\over dx}$$($$tan^{-1}x$$ $$\int$$ x dx} dx

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

$$\implies$$ I = $$x^2\over 2$$$$tan^{-1}x$$ – $$1\over 2$$ $$\int$$ $$x^2 + 1 – 1\over x^2 + 1$$ dx

$$\implies$$ I = $$x^2\over 2$$$$tan^{-1}x$$ – $$1\over 2$$ $$\int$$ 1 – $${1\over x^2 + 1}$$ dx

$$\implies$$ I = $$x^2\over 2$$$$tan^{-1}x$$ – $$1\over 2$$ ($$x – tan^{-1}x$$) + C

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