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Question Number 89717 by abdomathmax last updated on 18/Apr/20
$${let}\:\:{f}\left({x}\right)=\mathrm{2}{x}−\sqrt{{x}−\mathrm{1}} \\ $$$${find}\:\int\:\:\:\frac{{f}^{−\mathrm{1}} \left({x}\right)}{{f}\left({x}\right)}{dx}\:\: \\ $$
Answered by MJS last updated on 19/Apr/20
$${y}=\mathrm{2}{x}−\sqrt{{x}−\mathrm{1}} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\geqslant\mathrm{1} \\ $$$$\mathrm{range}\:{y}\geqslant\frac{\mathrm{17}}{\mathrm{16}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\begin{cases}{{y}=\frac{\mathrm{4}{x}+\mathrm{1}−\sqrt{\mathrm{8}{x}−\mathrm{15}}}{\mathrm{8}};\:\frac{\mathrm{15}}{\mathrm{8}}\leqslant{x}\leqslant\mathrm{2}}\\{{y}=\frac{\mathrm{4}{x}+\mathrm{1}+\sqrt{\mathrm{8}{x}−\mathrm{15}}}{\mathrm{8}};\:{x}\geqslant\frac{\mathrm{15}}{\mathrm{8}}}\end{cases} \\ $$$$ \\ $$$$\int\frac{{f}^{−\mathrm{1}} \left({x}\right)}{{f}\left({x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{4}{x}+\mathrm{1}}{\mathrm{2}{x}−\sqrt{{x}−\mathrm{1}}}{dx}\pm\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\sqrt{\mathrm{8}{x}−\mathrm{15}}}{\mathrm{2}{x}−\sqrt{{x}−\mathrm{1}}}{dx} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{first}\:\mathrm{one}\:\mathrm{but}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{one}\:\mathrm{takes}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{of}\:\mathrm{time}\:\mathrm{and}\:\mathrm{space}... \\ $$
Commented by mathmax by abdo last updated on 19/Apr/20
$${thank}\:{you}\:{sir} \\ $$