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Question Number 99462 by mathmax by abdo last updated on 21/Jun/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+2\mathrm{x}-5$$$$1)\mathrm{determine}{\mathrm{f}}^{-1}\left(\mathrm{x}\right)$$$$2)\mathrm{find}\int \frac{{\mathrm{f}}^{-1}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\mathrm{dx}$$$$3)\mathrm{let}\mathrm{u}\left(\mathrm{x}\right){=}^3\sqrt{\mathrm{x}}+2\mathrm{find}\int \frac{{\mathrm{uof}}^{-1}\left(\mathrm{x}\right)}{\mathrm{uof}\left(\mathrm{x}\right)}\mathrm{dx}$$

Answered by 1549442205 last updated on 22/Jun/20

$$\mathrm{Putting}{\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\mathrm{u}\Rightarrow {\mathrm{fof}}^{-1}\left(\mathrm{x}\right)=\mathrm{x}\Rightarrow \mathrm{f}\left(\mathrm{u}\right)=\mathrm{x}$$$$\Rightarrow {\mathrm{u}}^3+2\mathrm{u}-5-\mathrm{x}=0\left(1\right).\mathrm{Setting}\mathrm{u}=-(\mathrm{m}+\mathrm{n})$$$$\Rightarrow \mathrm{m}+\mathrm{n}+\mathrm{u}=0\Rightarrow {\mathrm{u}}^3+{\mathrm{m}}^3+{\mathrm{n}}^3-3\mathrm{nmn}=0\left(2\right)$$$$\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get}\{\begin{array}{l}{\mathrm{m}}^3+{\mathrm{n}}^3=-5-\mathrm{x}\\ \mathrm{mn}=\frac{-2}{3}\end{array}$$$$\Rightarrow {({\mathrm{m}}^3-{\mathrm{n}}^3)}^2={({\mathrm{m}}^3+{\mathrm{n}}^3)}^2-{4\mathrm{m}}^3{\mathrm{n}}^3={\mathrm{x}}^2+10\mathrm{x}+\frac{707}{27}$$$$\Rightarrow \{\begin{array}{l}{\mathrm{m}}^3+{\mathrm{n}}^3=-5-\mathrm{x}\\ {\mathrm{m}}^3-{\mathrm{n}}^3=\sqrt{{\mathrm{x}}^2+10\mathrm{x}+\frac{707}{27}}\end{array}$$$$\Rightarrow \{\begin{array}{l}{\mathrm{m}}^3=\frac{-5-\mathrm{x}+\sqrt{{\mathrm{x}}^2+10\mathrm{x}+\frac{707}{27}}}{2}\\ {\mathrm{n}}^3=\frac{-5-\mathrm{x}-\sqrt{{\mathrm{x}}^2+10\mathrm{x}+\frac{707}{27}}}{2}\end{array}$$$$\Rightarrow {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\mathrm{u}=-(\mathrm{m}+\mathrm{n}){=}^3\sqrt{\frac{5+\mathrm{x}-\sqrt{{\mathrm{x}}^2+10\mathrm{x}+\frac{707}{27}}}{2}}+{}^3\sqrt{\frac{5+\mathrm{x}+\sqrt{{\mathrm{x}}^2+10\mathrm{x}+\frac{707}{27}}}{2}}$$$$\mathrm{b}/\int \frac{{\mathrm{f}}^{-1}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\mathrm{dx}.\mathrm{Putting}{\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\mathrm{u}\Rightarrow \mathrm{f}\left(\mathrm{u}\right)=\mathrm{x}\Rightarrow \mathrm{dx}=\mathrm{f}{}^{\prime }\left(\mathrm{u}\right)\mathrm{du}=({3\mathrm{u}}^2+2)\mathrm{du}$$$$\mathrm{and}\mathrm{x}={\mathrm{u}}^3+2\mathrm{u}-5\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={({\mathrm{u}}^3+2\mathrm{u}-5)}^3+{2\mathrm{u}}^3+4\mathrm{u}-15$$$$={\mathrm{u}}^9+{8\mathrm{u}}^3-125+{6\mathrm{u}}^7-{15\mathrm{u}}^6-{60\mathrm{u}}^2+{12\mathrm{u}}^5+{75\mathrm{u}}^3$$$$+150\mathrm{u}-{60\mathrm{u}}^4+{2\mathrm{u}}^3+4\mathrm{u}-15=$$$${\mathrm{u}}^9+{6\mathrm{u}}^7-{15\mathrm{u}}^6+{12\mathrm{u}}^5-{60\mathrm{u}}^4+{77\mathrm{u}}^3-{60\mathrm{u}}^2+154\mathrm{u}-140$$$$\mathrm{Hence},\mathrm{F}=\int \frac{\mathrm{u}({3\mathrm{u}}^2+2)\mathrm{du}}{{({\mathrm{u}}^3+2\mathrm{u}-5)}^3+2({\mathrm{u}}^3+2\mathrm{u}-5)-5}$$

Commented by mathmax by abdo last updated on 21/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

Commented by 1549442205 last updated on 25/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{wellcome},\mathrm{sir}.$$

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