If \(\int \frac{dx}{{{\left( {{x}^{2}}-2x+10 \right)}^{2}}}=A\left( ta{{n}^{-1}}\left( \frac{x-1}{3} \right)+\frac{f\left( x \right)}{{{x}^{2}}-2x+10} \right)+C\) where C is a constant of integration, then:

\(A = \frac{1}{{81}}{\rm{\;and\;}}f\left( x \right) = 3\left( {x - 1} \right)\)

\(A = \frac{1}{{54}}{\rm{\;and\;}}f\left( x \right) = 3\left( {x - 1} \right)\)

\(A = \frac{1}{{27}}{\rm{\;and\;}}f\left( x \right) = 9\left( {x - 1} \right)\)

\(A = \frac{1}{{54}}{\rm{\;and\;}}f\left( x \right) = 9{(x - 1)^2}\)

If \(\int \frac{dx}{{{\left( {{x}^{2}}-2x+10 \right)}^{2}}}=A\left( t

1.	If \(\int \frac{dx}{{{\left( {{x}^{2}}-2x+10 \right)}^{2}}}=A\left( ta{{n}^{-1}}\left( \frac{x-1}{3} \right)+\frac{f\left( x \right)}{{{x}^{2}}-2x+10} \right)+C\) where C is a constant of integration, then:
A.	\(A = \frac{1}{{54}}{\rm{\;and\;}}f\left( x \right) = 3\left( {x - 1} \right)\)
B.	\(A = \frac{1}{{81}}{\rm{\;and\;}}f\left( x \right) = 3\left( {x - 1} \right)\)
C.	\(A = \frac{1}{{27}}{\rm{\;and\;}}f\left( x \right) = 9\left( {x - 1} \right)\)
D.	\(A = \frac{1}{{54}}{\rm{\;and\;}}f\left( x \right) = 9{(x - 1)^2}\)
Answer» B. \(A = \frac{1}{{81}}{\rm{\;and\;}}f\left( x \right) = 3\left( {x - 1} \right)\)

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If \(\int \frac{dx}{{{\left( {{x}^{2}}-2x+10 \right)}^{2}}}=A\left( ta{{n}^{-1}}\left( \frac{x-1}{3} \right)+\frac{f\left( x \right)}{{{x}^{2}}-2x+10} \right)+C\) where C is a constant of integration, then:

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