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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} \mathrm{F}(x) & =(\mathrm{fof})(x) \ & =\mathrm{f}(\mathrm{f}(x)) \ & =\mathrm{f}\left(\frac{x}{x+1}ight) \ & =\frac{\frac{x}{x+1}}{\frac{x}{x+1}+1} \ & =\frac{x}{2 x+1}\end{aligned}$ $\begin{aligned} \int \mathrm{F}(x) \mathrm{d} x & =\int \frac{x}{2 x+1} \mathrm{~d} x \ & =\int \frac{\frac{1}{2}(2 x+1)-\frac{2}{4}}{2 x+1} \mathrm{~d} x \ & =\frac{1}{2} \int \mathrm{~d} x-\frac{1}{4} \int \frac{2}{2 x+1} \mathrm{~d} x \ & =\frac{1}{2} x-\frac{1}{4} \log |2 x+1|+c\end{aligned}$

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

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