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Question Number 183806 by cortano1 last updated on 30/Dec/22

$$\int \frac{\sqrt{x+\sqrt{x^2+25}}}{x}dx=?$$

Answered by Ar Brandon last updated on 30/Dec/22

$$I=\int \frac{\sqrt{x+\sqrt{x^2+25}}}{x}dx,x=5\mathrm{sh}\vartheta$$$$=\int \frac{\sqrt{5\mathrm{sh}\vartheta +5\mathrm{ch}\vartheta }}{5\mathrm{sh}\vartheta }\left(5\mathrm{ch}\vartheta d\vartheta \right)=\sqrt{5}\int \left(\frac{e^{2\vartheta }+1}{e^{2\vartheta }-1}\right)e^{\frac{\vartheta }{2}}d\vartheta ,\vartheta =2\varphi$$$$=2\sqrt{5}\int \left(\frac{e^{4\varphi }+1}{e^{4\varphi }-1}\right)e^{\varphi }d\varphi =2\sqrt{5}\int \left(\frac{t^4+1}{t^4-1}\right)dt=2\sqrt{5}\int (1+\frac{2}{t^4-1})dt,t=e^{\varphi }$$$$=2\sqrt{5}\int (1+\frac{1}{t^2-1}-\frac{1}{t^2+1})dt=2\sqrt{5}(t+\frac{1}{2}\ln \mid \frac{t-1}{t+1}\mid -\arctan \left(t\right))+C$$$$=2\sqrt{5}(e^{\frac{\vartheta }{2}}+\frac{1}{2}\ln \mid e^{\frac{\vartheta }{2}}-1\mid -\frac{1}{2}\ln \mid e^{\frac{\vartheta }{2}}+1\mid -\arctan \left(e^{\frac{\vartheta }{2}}\right))+C$$$$=2\sqrt{5}\sqrt{x+\sqrt{x^2+25}}+\sqrt{5}\ln \mid \frac{\sqrt{x+\sqrt{x^2+25}}-1}{\sqrt{x+\sqrt{x^2+25}}+1}\mid -2\sqrt{5}\arctan \left(\sqrt{x+\sqrt{x^2+25}}\right)+C$$

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