If g(1) = g(2),then int_1^2 {{{left[ {f(g(x)) | vedclass

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(C) Let the integral be $I = \int_1^2 \frac{f'(g(x)) \cdot g'(x)}{f(g(x))} \; dx$. Substitute $u = f(g(x))$. Then,the differential is $du = f'(g(x)) \

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$0$

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(C) Let the integral be $I = \int_1^2 \frac{f'(g(x)) \cdot g'(x)}{f(g(x))} \; dx$. Substitute $u = f(g(x))$. Then,the differential is $du = f'(g(x)) \cdot g'(x) \; dx$. Now,change the limits of integration: When $x = 1$,$u = f(g(1))$. When $x = 2$,$u = f(g(2))$. Since $g(1) = g(2)$,it follows that $f(g(1)) = f(g(2))$. Therefore,the integral becomes $I = \int_{f(g(1))}^{f(g(1))} \frac{1}{u} \; du$. Since the lower and upper limits of the definite integral are identical,the value of the integral is $0$.

If $g(1) = g(2)$,then $\int_1^2 {{{\left[ {f(g(x))} \right]}^{ - 1}}} f'\{ g(x)\} \;g'(x)\;dx$ is equal to

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