If g(1) = g(2),then the value of int_{1}^{2} | vedclass

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(C) Let $I = \int_{1}^{2} [f\{g(x)\}]^{-1} f'\{g(x)\} g'(x) dx$.Substitute $u = g(x)$,then $du = g'(x) dx$.When $x = 1$,$u = g(1)$.When $x = 2$,$u = g

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(C) Let $I = \int_{1}^{2} [f\{g(x)\}]^{-1} f'\{g(x)\} g'(x) dx$.Substitute $u = g(x)$,then $du = g'(x) dx$.When $x = 1$,$u = g(1)$.When $x = 2$,$u = g(2)$.Since $g(1) = g(2) = t$,the integral becomes:$I = \int_{g(1)}^{g(2)} [f(u)]^{-1} f'(u) du$$I = \int_{t}^{t} \frac{f'(u)}{f(u)} du$By the property of definite integrals,if the upper and lower limits are equal,the integral is $0$.$I = [\ln|f(u)|]_{t}^{t} = \ln|f(t)| - \ln|f(t)| = 0$.

If $g(1) = g(2)$,then the value of $\int_{1}^{2} [f\{g(x)\}]^{-1} f'\{g(x)\} g'(x) dx$ is:

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