If f(frac{x - 1}{x + 1}) = x + 1, then int f( | vedclass

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(B) Let $t = \frac{x - 1}{x + 1}$.Then $t(x + 1) = x - 1 \implies tx + t = x - 1 \implies x(t - 1) = -1 - t \implies x = \frac{t + 1}{1 - t}$.Substitu

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$-2 \log |1 - x| + c$

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(B) Let $t = \frac{x - 1}{x + 1}$.Then $t(x + 1) = x - 1 \implies tx + t = x - 1 \implies x(t - 1) = -1 - t \implies x = \frac{t + 1}{1 - t}$.Substitute $x$ into the given function: $f(t) = \frac{t + 1}{1 - t} + 1 = \frac{t + 1 + 1 - t}{1 - t} = \frac{2}{1 - t}$.Now, $\int f(x) dx = \int \frac{2}{1 - x} dx$.Let $u = 1 - x$, then $du = -dx$, so $dx = -du$.$\int \frac{2}{u} (-du) = -2 \int \frac{1}{u} du = -2 \log |u| + c = -2 \log |1 - x| + c$.

If $f(\frac{x - 1}{x + 1}) = x + 1$, then $\int f(x) dx = \dots$

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