If ∫dx(x2−2x+10)2=A(tan⁡−1(x−13)+f(x)x2−2x+10)+C,\int \frac{dx}{(x^2 - 2x + 10)^2} = A \left( \tan^{-1} \left( \frac{x-1}{3} \right) + \frac{f(x)}{x^2 - 2x + 10} \right) + C,…

Mathematics

Problem

If ∫dx(x2−2x+10)2=A(tan⁡−1(x−13)+f(x)x2−2x+10)+C,\int \frac{dx}{(x^2 - 2x + 10)^2} = A \left( \tan^{-1} \left( \frac{x-1}{3} \right) + \frac{f(x)}{x^2 - 2x + 10} \right) + C, where CC is a constant of integration, then (a) A=154A = \frac{1}{54} and f(x)=3(x−1)f(x) = 3(x-1) (b) A=127A = \frac{1}{27} and f(x)=9(x−1)f(x) = 9(x-1) (c) A=181A = \frac{1}{81} and f(x)=3(x−1)f(x) = 3(x-1) (d) A=154A = \frac{1}{54} and f(x)=9(x−1)2f(x) = 9(x-1)^2

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