If f(x) = frac{x}{x+1}, x neq -1 and (fof)(x) | vedclass

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(B) Given $f(x) = \frac{x}{x+1}$.We find $F(x) = (fof)(x) = f(f(x)) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}

Vedclass · Accepted Answer

$\frac{x}{2} - \frac{1}{4} \log |2x+1| + c$,where $c$ is a constant of integration.

Vedclass · Answer

(B) Given $f(x) = \frac{x}{x+1}$.We find $F(x) = (fof)(x) = f(f(x)) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + (x+1)} = \frac{x}{2x+1}$.Now,we evaluate the integral $\int F(x) \, dx = \int \frac{x}{2x+1} \, dx$.To integrate,we manipulate the numerator: $\frac{x}{2x+1} = \frac{1}{2} \left( \frac{2x+1-1}{2x+1} \right) = \frac{1}{2} \left( 1 - \frac{1}{2x+1} \right) = \frac{1}{2} - \frac{1}{2(2x+1)}$.Integrating term by term: $\int \left( \frac{1}{2} - \frac{1}{2(2x+1)} \right) \, dx = \frac{1}{2}x - \frac{1}{2} \cdot \frac{1}{2} \log |2x+1| + c = \frac{x}{2} - \frac{1}{4} \log |2x+1| + c$.

If $f(x) = \frac{x}{x+1}, x \neq -1$ and $(fof)(x) = F(x)$,then $\int F(x) \, dx$ is

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