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Exercise 4.2 · Q19

Q.Evaluate: ∫01xtan⁡−1x dx\int_0^1 x\tan^{-1}x\,dx

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u=tan⁡−1x, dv=x dx⇒v=x22u=\tan^{-1}x,\ dv=x\,dx\Rightarrow v=\frac{x^2}2.

∫xtan⁡−1x dx=x22tan⁡−1x−12∫x21+x2 dx=x22tan⁡−1x−12(x−tan⁡−1x).\int x\tan^{-1}x\,dx=\frac{x^2}2\tan^{-1}x-\frac12\int\frac{x^2}{1+x^2}\,dx=\frac{x^2}2\tan^{-1}x-\frac12\Big(x-\tan^{-1}x\Big). …

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