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Question Number 110620 by peter frank last updated on 29/Aug/20

Answered by Dwaipayan Shikari last updated on 29/Aug/20

$$f\left(x\right)={tan}^{-1}x$$$$f^{\prime }\left(x\right)=\frac{1}{1+x^2}={(1+x^2)}^{-1}$$$$f^{\prime }\left(x\right)=1-x^2+x^4-....$$$$\int f^{\prime }\left(x\right)=\int 1-x^2+x^4-...$$$$f\left(x\right)=x-\frac{x^3}{3}+\frac{x^5}{5}-....+C(C=0)$$$${tan}^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-.....$$$$$$

Commented by peter frank last updated on 30/Aug/20

$$\mathrm{thank}\mathrm{you}$$

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