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Question Number 208235 by efronzo1 last updated on 08/Jun/24

Answered by som(math1967) last updated on 08/Jun/24

$$heref\left(x\right)=f^{-1}\left(x\right)$$$$\underset{2}{\stackrel{4}{\int }}\frac{1-x}{1+x}dx$$$$=\underset{2}{\stackrel{4}{\int }}\frac{2dx}{1+x}-\underset{2}{\stackrel{4}{\int }}\frac{1+x}{1+x}dx$$$$={[2ln(1+x)-x]}_2^4$$$$=2ln5-4-2ln3+2$$$$=2ln\frac{5}{3}-2$$

Answered by shunmisaki007 last updated on 08/Jun/24

$$\mathrm{Let}y=f^{-1}\left(x\right)$$$$f\left(y\right)=x=\frac{1-y}{1+y}$$$$x(1+y)=1-y$$$$x+xy=1-y$$$$xy+y=1-x$$$$y(1+x)=1-x$$$$\Rightarrow y=f^{-1}\left(x\right)=\frac{1-x}{1+x}$$$$\underset{2}{\stackrel{4}{\int }}{f}^{-1}\left(x\right)dx=\underset{2}{\stackrel{4}{\int }}\frac{1-x}{1+x}dx=\underset{2}{\stackrel{4}{\int }}\frac{1-x+1-1}{1+x}dx$$$$=\underset{2}{\stackrel{4}{\int }}\frac{2-x-1}{1+x}dx=\underset{2}{\stackrel{4}{\int }}\frac{2-(x+1)}{1+x}dx$$$$=\underset{2}{\stackrel{4}{\int }}\frac{2}{1+x}dx-\underset{2}{\stackrel{4}{\int }}\frac{1+x}{1+x}dx$$$$={\left[2\ln (1+x)\right]}_2^4-{\left[x\right]}_2^4$$$$=(2\ln (1+4)-2\ln (1+2))-(4-2)$$$$=2(\ln \left(5\right)-\ln \left(3\right))-2$$$$\therefore \underset{2}{\stackrel{4}{\int }}{f}^{-1}\left(x\right)dx=2\ln \left(\frac{5}{3}\right)-2\approx -0.987★$$$$$$

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