If int f(x) dx = g(x) + c, then int f^{-1}(x) | vedclass

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(C) Let $y = f^{-1}(x)$, then $x = f(y)$.Differentiating both sides with respect to $y$, we get $dx = f'(y) dy$.Now, $\int f^{-1}(x) dx = \int y f'(y)

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$x f^{-1}(x) - g(f^{-1}(x)) + c$

Vedclass · Answer

(C) Let $y = f^{-1}(x)$, then $x = f(y)$.Differentiating both sides with respect to $y$, we get $dx = f'(y) dy$.Now, $\int f^{-1}(x) dx = \int y f'(y) dy$.Using integration by parts: $\int u dv = uv - \int v du$, where $u = y$ and $dv = f'(y) dy$.Then $du = dy$ and $v = f(y)$.So, $\int y f'(y) dy = y f(y) - \int f(y) dy$.Since $\int f(y) dy = g(y) + c$, we have $y f(y) - g(y) + c$.Substituting $y = f^{-1}(x)$, we get $f^{-1}(x) \cdot f(f^{-1}(x)) - g(f^{-1}(x)) + c$.Since $f(f^{-1}(x)) = x$, the result is $x f^{-1}(x) - g(f^{-1}(x)) + c$.

If $\int f(x) dx = g(x) + c$, then $\int f^{-1}(x) dx =$

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