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Question Number 27081 by abdo imad last updated on 01/Jan/18

$$letgivef\left(x\right)=\frac{x}{4x^2-1}find{f}^{\left(n\right)}\left(x\right).$$

Answered by prakash jain last updated on 01/Jan/18

$$f\left(x\right)\frac{x}{4x^2-1}=\frac{1}{4}(\frac{1}{2x-1}+\frac{1}{2x+1})$$$$f^{\left(n\right)}\left(x\right)=\frac{1}{4}(\frac{{(-2)}^{n}n!}{{(2x-1)}^{n+1}}+\frac{{(-2)}^{n}n!}{{(2x+1)}^{n+1}})$$

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