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Question Number 41586 by MJS last updated on 09/Aug/18

$$f\left(x\right)=\sqrt{-3+\sqrt{\frac{x+1}{x-1}}}$$$$\int f\left(x\right)=?$$$$\int f^{-1}\left(x\right)=?$$

Commented by prof Abdo imad last updated on 10/Aug/18

$$f\left(x\right)=y\iff x=f^{-1}\left(y\right)\Rightarrow \sqrt{-3+\sqrt{\frac{x+1}{x-1}}}=y\Rightarrow$$$$\sqrt{\frac{x+1}{x-1}}=y^2+3\Rightarrow \frac{x+1}{x-1}={(y^2+3)}^2\Rightarrow$$$$x+1=x{(y^2+3)}^2-{(y^2+3)}^2\Rightarrow$$$$(1-{(y^2+3)}^2)x=-1-{(y^2+3)}^2\Rightarrow$$$$x=\frac{{(y^2+3)}^2+1}{{(y^2+3)}^2-1}=1+\frac{2}{{(y^2+3)}^2-1}\Rightarrow$$$$f^{-1}\left(x\right)=1+\frac{2}{{(x^2+3)}^2-1}\Rightarrow$$$$\int f^{-1}\left(x\right)dx=x+2\int \frac{dx}{{(x^2+3)}^2-1}+cbut$$$$\int \frac{2dx}{{(x^2+3)}^2-1}=\int \frac{2dx}{(x^2+4)(x^2+2)}$$$$=\int \{\frac{1}{x^2+2}-\frac{1}{x^2+4}\}dx=\int \frac{dx}{x^2+2}-\int \frac{dx}{x^2+4}$$$$\int \frac{dx}{x^2+2}{=}_{x=\sqrt{2}t}\int \frac{\sqrt{2}dt}{2(1+t^2)}=\frac{1}{\sqrt{2}}arctan\left(\frac{x}{\sqrt{2}}\right)$$$$\int \frac{dx}{x^2+4}{=}_{x=2t}\int \frac{2dt}{4(1+t^2)}=\frac{1}{2}arctan\left(\frac{t}{2}\right)\Rightarrow$$$$\int f^{-1}\left(x\right)dx=x+\frac{1}{\sqrt{2}}arctan\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{2}arctan\left(\frac{x}{2}\right)+c$$

Answered by MJS last updated on 09/Aug/18

$$\int \sqrt{-3+\sqrt{\frac{x+1}{x-1}}}dx=$$$$[t=\sqrt{-3+\sqrt{\frac{x+1}{x-1}}}\to dx=-\frac{8t(t^2+3)}{{(t^2+2)}^2{(t^2+4)}^2}dt]$$$$=-8\int \frac{t^2(t^2+3)}{{(t^2+2)}^2{(t^2+4)}^2}dt=$$$$=-8\int (-\frac{1}{2{(t^2+2)}^2}+\frac{1}{{(t^2+4)}^2}+\frac{1}{4(t^2+2)}-\frac{1}{4(t^2+4)})dt=$$$$=4\int \frac{dt}{{(t^2+2)}^2}-8\int \frac{dt}{{(t^2+4)}^2}-2\int \frac{dt}{t^2+2}+2\int \frac{dt}{t^2+4}=$$$$[\int \frac{dx}{{(x^2+p)}^2}=\frac{x}{2p(x^2+p)}+\frac{\sqrt{p^3}}{2p^3}\arctan \frac{x\sqrt{p}}{p}]$$$$[\int \frac{dx}{x^2+p}=\frac{\sqrt{p}}{p}\arctan \frac{x\sqrt{p}}{p}]$$$$=\frac{t}{t^2+2}+\frac{\sqrt{2}}{2}\arctan \frac{t\sqrt{2}}{2}-\frac{t}{t^2+4}-\frac{1}{2}\arctan \frac{t}{2}-\sqrt{2}\arctan \frac{t\sqrt{2}}{2}+\arctan \frac{t}{2}=$$$$=\frac{2t}{(t^2+2)(t^2+4)}+\frac{1}{2}(\arctan \frac{t}{2}-\sqrt{2}\arctan \frac{t\sqrt{2}}{2})=$$$$=\sqrt{(\sqrt{x+1}-3\sqrt{x-1})\sqrt{{(x-1)}^3}}+\frac{1}{2}(\arctan \left(\frac{1}{2}\sqrt{-3+\sqrt{\frac{x+1}{x-1}}}\right)-\sqrt{2}\arctan \left(\frac{1}{2}\sqrt{-6+2\sqrt{\frac{x+1}{x-1}}}\right))+C$$

Answered by ajfour last updated on 10/Aug/18

$${(y^2+3)}^2=1+\frac{2}{x-1}$$$$x=1+\frac{2}{{(y^2+3)}^2-1}$$$$f^{-1}\left(x\right)=1+\frac{2}{{(x^2+3)}^2-1}$$$$I=\int [1+\frac{2}{{(x^2+3)}^2-1}]dx$$$$=x+2I_1$$$$I_1=\int \frac{dx}{{(x^2+3)}^2-1}=\int \frac{dx}{(x^2+2)(x^2+4)}$$$$=\frac{1}{2}\int \frac{dx}{x^2+2}-\frac{1}{2}\int \frac{dx}{x^2+4}$$$$hence$$$$\int f^{-1}\left(x\right)dx=x+\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}$$$$-\frac{1}{2}{\tan }^{-1}\frac{x}{2}+c.$$

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