If int frac{dx}{(x^2 - 2x + 10)^2} = A left( | vedclass

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(C) Let $I = \int \frac{dx}{(x^2 - 2x + 10)^2} = \int \frac{dx}{((x - 1)^2 + 9)^2}$.Substitute $x - 1 = 3 	an 	heta$,so $dx = 3 \sec^2 	heta d	het

Vedclass · Accepted Answer

$A = \frac{1}{54}$ and $f(x) = 3(x - 1)$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(x^2 - 2x + 10)^2} = \int \frac{dx}{((x - 1)^2 + 9)^2}$.Substitute $x - 1 = 3 	an 	heta$,so $dx = 3 \sec^2 	heta d	heta$.The integral becomes $\int \frac{3 \sec^2 	heta d	heta}{(9 	an^2 	heta + 9)^2} = \int \frac{3 \sec^2 	heta d	heta}{81 \sec^4 	heta} = \frac{1}{27} \int \cos^2 	heta d	heta$.Using $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get $\frac{1}{54} \int (1 + \cos 2	heta) d	heta = \frac{1}{54} (	heta + \frac{\sin 2	heta}{2}) + C$.Since $	an 	heta = \frac{x - 1}{3}$,then $	heta = 	an^{-1} \left( \frac{x - 1}{3} ight)$.Also,$\sin 2	heta = 2 \sin 	heta \cos 	heta = 2 \left( \frac{x - 1}{\sqrt{(x - 1)^2 + 9}} ight) \left( \frac{3}{\sqrt{(x - 1)^2 + 9}} ight) = \frac{6(x - 1)}{x^2 - 2x + 10}$.Substituting these back,$I = \frac{1}{54} \left( 	an^{-1} \left( \frac{x - 1}{3} ight) + \frac{3(x - 1)}{x^2 - 2x + 10} ight) + C$.Comparing with the given form,$A = \frac{1}{54}$ and $f(x) = 3(x - 1)$.

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

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