# What is the integration of tan inverse root x ?

## Solution :

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

By Applying integration by parts,

Taking $$tan^{-1}\sqrt{x}$$ as first function and 1 as second function. Then

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

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

Let $$\sqrt{x}$$ = t

$$1\over 2\sqrt{x}$$ dx = dt $$\implies$$ dx = 2t dt

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

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

$$\implies$$ I = x$$tan^{-1}\sqrt{x}$$ – t + $$tan^{-1}t$$ + C

Hence, I = x$$tan^{-1}\sqrt{x}$$ – $$\sqrt{x}$$ + $$tan^{-1}\sqrt{x}$$ + C

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