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Question Number 96092 by mhmd last updated on 29/May/20

$$find\int \frac{dx}{{tan}^{-1}\left(x\right)}$$

Commented by bemath last updated on 30/May/20

$$\int {\cot }^{-1}\left(x\right)dx=\mathrm{I}$$$$\mathrm{by}\mathrm{parts}.\mathrm{u}={\cot }^{-1}\left(x\right)\Rightarrow du=-\sqrt{1-x^2}dx$$$$dv=dx\Rightarrow v=x$$$$\mathrm{I}=x{\cot }^{-1}\left(x\right)+\int x\sqrt{1-x^2}dx$$$$\mathrm{I}=x{\cot }^{-1}\left(x\right)-\frac{1}{2}\int \sqrt{1-x^2}d(1-x^2)$$$$\mathrm{I}=x{\cot }^{-1}\left(x\right)-\frac{1}{3}\sqrt{{(1-x^2)}^3}+c$$

Commented by Kunal12588 last updated on 30/May/20

$$\frac{1}{{tan}^{-1}x}\ne {cot}^{-1}x$$$$[{tan}^{-1}\left(\frac{1}{x}\right)={cot}^{-1}x]$$

Commented by mathmax by abdo last updated on 30/May/20

$$\mathrm{answer}\mathrm{not}\mathrm{correct}!$$

Commented by MJS last updated on 30/May/20

$$\mathrm{this}\mathrm{cannot}\mathrm{be}\mathrm{solved}$$

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