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Question Number 113628 by mathmax by abdo last updated on 14/Sep/20

$$\mathrm{find}\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2-1}+(\mathrm{x}-1)\sqrt{{\mathrm{x}}^2+1}}$$

Answered by MJS_new last updated on 16/Sep/20

$$\int \frac{dx}{(x+1)\sqrt{x^2-1}+(x-1)\sqrt{x^2+1}}=$$$$=\int \frac{(x+1)\sqrt{x^2-1}-(x-1)\sqrt{x^2+1}}{2(x-1)(2x^2+x+1)}dx=$$$$=\frac{1}{4}\int \frac{\sqrt{x^2-1}}{x-1}dx-\frac{1}{2}\int \frac{\sqrt{x^2+1}}{2x^2+x+1}dx-\frac{1}{4}\int \frac{(2x+1)\sqrt{x^2-1}}{2x^2+x+1}dx$$$$$$$$\frac{1}{4}\int \frac{\sqrt{x^2-1}}{x-1}dx=\left(\right)$$$$[t=x+\sqrt{x^2-1}\to dx=\frac{\sqrt{x^2-1}}{x+\sqrt{x^2-1}}dt]$$$$=\frac{1}{8}\int \frac{{(t+1)}^2}{t^2}dt=\frac{t^2-1}{8t}+\frac{1}{4}\ln t=$$$$=\frac{1}{4}\sqrt{x^2-1}+\frac{1}{4}\ln (x+\sqrt{x^2-1})$$$$$$$$-\frac{1}{2}\int \frac{\sqrt{x^2+1}}{2x^2+x+1}dx=$$$$[t=x+\sqrt{x^2+1}\to dx=\frac{\sqrt{x^2+1}}{x+\sqrt{x^2+1}}]$$$$=-\frac{1}{4}\int \frac{{(t^2+1)}^2}{t(t^4+t^3-t+1)}dt$$$$\mathrm{now}\mathrm{we}\mathrm{must}\mathrm{decompose}$$$$t^4+t^3-t+1=(t^2+\alpha t+\beta )(t^2+\gamma t+\delta )$$$$\alpha =\frac{1+\sqrt{5+4\sqrt{2}}}{2}$$$$\beta =\frac{1+\sqrt{2}}{2}+\frac{\sqrt{5+4\sqrt{2}}+\sqrt{-35+28\sqrt{2}}}{8}$$$$\gamma =\frac{1-\sqrt{5+4\sqrt{2}}}{2}$$$$\delta =\frac{1+\sqrt{2}}{2}-\frac{\sqrt{5+4\sqrt{2}}+\sqrt{-35+28\sqrt{2}}}{8}$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{too}\mathrm{tired}\mathrm{now}$$$$$$$$-\frac{1}{4}\int \frac{(2x+1)\sqrt{x^2-1}}{2x^2+x+1}dx=$$$$[t=x+\sqrt{x^2-1}\to dx=\frac{\sqrt{x^2-1}}{x+\sqrt{x^2-1}}]$$$$=-\frac{1}{8}\int \frac{{(t^2-1)}^2(t^2+t+1)}{t^2(t^4+t^3+4t^2+t+1)}dt$$$$\mathrm{again}\mathrm{to}\mathrm{decompose}$$$$t^4+t^3+4t^2+t+1=(t^2+\mathfrak{a}t+\mathfrak{b})(t^2+\mathfrak{c}t+\mathfrak{d})$$$$\mathfrak{a}=\frac{1+\sqrt{-11+8\sqrt{2}}}{2}$$$$\mathfrak{b}=\frac{1+2\sqrt{2}}{2}+\frac{\sqrt{-11+8\sqrt{2}}+\sqrt{77+56\sqrt{2}}}{8}$$$$\mathfrak{c}=\frac{1-\sqrt{-11+8\sqrt{2}}}{2}$$$$\mathfrak{d}=\frac{1+2\sqrt{2}}{2}-\frac{\sqrt{-11+8\sqrt{2}}+\sqrt{77+56\sqrt{2}}}{8}$$$$\mathrm{please}\mathrm{continue}\mathrm{if}\mathrm{you}\mathrm{need}\mathrm{it},\mathrm{I}\mathrm{go}\mathrm{to}\mathrm{bed}...$$

Commented by Dwaipayan Shikari last updated on 16/Sep/20

$$\frac{1}{4}\int \sqrt{\frac{x+1}{x-1}}$$$$\frac{1}{4}\int \frac{x+1}{\sqrt{x^2-1}}=\frac{1}{8}\int \frac{2x}{\sqrt{x^2-1}}+\frac{1}{4}\int \frac{1}{\sqrt{x^2-1}}=\frac{1}{4}\sqrt{x^2-1}+\frac{1}{4}log(x+\sqrt{x^2-1})$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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