A) \[x{{f}^{-1}}(x)+C\]
B) \[f({{g}^{-1}})(x))+C\]
C) \[x{{f}^{-1}}(x)-g({{f}^{-1}})(x))+C\]
D) \[{{g}^{-1}}(x)+C\]
Correct Answer: C
Solution :
[c] Let \[I=\int{{{f}^{-1}}(x)dx}\] and \[{{f}^{-1}}(x)=t\Rightarrow x=f(t)\Rightarrow dx=f'(t)dt\] Put value of \[dx\] and \[{{f}^{-1}}(x)\] in I, we get \[I=\int{tf'(t)dt}\] Now, integrate it by parts, \[I=tf(t)-\int{f(t)dt}\] Given, \[\int{f(x)dx=g(x)+C}\] \[\therefore I=tf(t)-[g(t)]+C\] Now, by putting value of t, f (t) and g(t) we get, \[I=x{{f}^{-1}}(x)-g[{{f}^{-1}}(x)]+C\]
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