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Question Number 137004 by Mathspace last updated on 28/Mar/21

$$find\int \frac{\sqrt{x}}{\sqrt{x-1}+\sqrt{x+1}}dx$$

Answered by aleks041103 last updated on 28/Mar/21

$$\frac{\sqrt{x}}{\sqrt{x-1}+\sqrt{x+1}}=$$$$=\frac{\sqrt{x}}{\sqrt{x-1}+\sqrt{x+1}}\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=$$$$=\sqrt{x}(\sqrt{x+1}-\sqrt{x-1})$$$$$$$$I\left(a\right)=\int \sqrt{x(x+2a)}dx=$$$$=\int \sqrt{((x+a)-a)((x+a)+a)}dx=$$$$=\int \sqrt{{(x+a)}^2-a^2}dx=$$$$=a^2\int \sqrt{{(1+x/a)}^2-1}d(1+\frac{x}{a})$$$$cosh\left(t\right)=1+\frac{x}{a}$$$$sinh\left(t\right)dt=d(1+\frac{x}{a})$$$$I\left(a\right)=a^2\int \sqrt{{cosh}^2\left(t\right)-1}sinh\left(t\right)dt=$$$$=a^2\int {sinh}^2\left(t\right)dt$$$${sinh}^2\left(t\right)=cosh\left(2t\right)-1$$$$I\left(a\right)=a^2(\frac{1}{2}sinh\left(2t\right)-t)$$$$t=arccosh(1+x/a)=ln(1+\frac{x}{a}+\sqrt{{(1+x/a)}^2-1})=$$$$=ln(1+a^{-1}x+a^{-1}\sqrt{x(x+2a)})$$$$\frac{1}{2}sinh\left(2t\right)=sinh\left(t\right)cosh\left(t\right)=$$$$=cosh\left(t\right)\sqrt{{cosh}^2\left(t\right)-1}=$$$$=(1+x/a)\sqrt{{(1+x/a)}^2-1}=$$$$=a^{-2}(x+a)\sqrt{x(x+2a)}$$$$I\left(a\right)=(x+a)\sqrt{x(x+2a)}-a^{2}ln(1+a^{-1}x+a^{-1}\sqrt{x(x+2a)})$$$$$$$$\int \frac{\sqrt{x}}{\sqrt{x-1}+\sqrt{x+1}}dx=\int \sqrt{x}(\sqrt{x+1}-\sqrt{x-1})dx=$$$$=I\left(1/2\right)-I(-1/2)=$$$$=\frac{1}{2}((2x+1)\sqrt{x(x+1)}-(2x-1)\sqrt{x(x-1)})-\frac{1}{4}ln\left(\frac{1+2x+2\sqrt{x(x+1)}}{1-2x-2\sqrt{x(x-1)}}\right)$$$$\int \frac{\sqrt{x}}{\sqrt{x-1}+\sqrt{x+1}}dx=$$$$=((2x+1)\sqrt{x(x+1)}-(2x-1)\sqrt{x(x-1)})-\frac{1}{4}ln\left(\frac{1+2x+2\sqrt{x(x+1)}}{1-2x-2\sqrt{x(x-1)}}\right)+C$$

Answered by mathmax by abdo last updated on 29/Mar/21

$$\mathrm{thankx}\mathrm{sir}.$$

Answered by EDWIN88 last updated on 29/Mar/21

$$\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{x}+1}}=\frac{\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}-1}-\sqrt{\mathrm{x}+1})}{(\mathrm{x}-1)-(\mathrm{x}+1)}$$$$=-\frac{\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}-1}-\sqrt{\mathrm{x}+1})}{2}$$$$\mathrm{E}=-\frac{1}{2}\int \sqrt{{\mathrm{x}}^2-\mathrm{x}}\mathrm{dx}+\frac{1}{2}\int \sqrt{{\mathrm{x}}^2+\mathrm{x}}\mathrm{dx}$$$$\mathrm{E}=-\frac{1}{2}\int \sqrt{{(\mathrm{x}-\frac{1}{2})}^2-\frac{1}{4}}\mathrm{dx}+\frac{1}{2}\int \sqrt{{(\mathrm{x}+\frac{1}{2})}^2-\frac{1}{4}}\mathrm{dx}$$$${\mathrm{E}}_1=-\frac{1}{2}\int \sqrt{{(\mathrm{x}-\frac{1}{2})}^2-\frac{1}{4}}\mathrm{dx}$$$$\mathrm{let}\mathrm{x}-\frac{1}{2}=\frac{1}{2}\sec \mathrm{t}\mathrm{dx}=\frac{1}{2}\sec \mathrm{t}\tan \mathrm{t}\mathrm{dt}$$$${\mathrm{E}}_1=-\frac{1}{2}\int \frac{1}{2}\tan \mathrm{t}.\frac{1}{2}\sec \mathrm{t}\tan \mathrm{t}\mathrm{dt}$$$${\mathrm{E}}_1=-\frac{1}{8}\int \sec \mathrm{t}(\sec {}^2\mathrm{t}-1)\mathrm{dt}$$$${\mathrm{E}}_1=-\frac{1}{8}\int \sec {}^3\mathrm{t}-\sec \mathrm{t}\mathrm{dt}$$$$\mathrm{similary}\mathrm{to}{\mathrm{E}}_2=\frac{1}{2}\int \sqrt{{(\mathrm{x}+\frac{1}{2})}^2-\frac{1}{4}}\mathrm{dx}$$

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