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Question Number 81921 by M±th+et£s last updated on 16/Feb/20

Answered by MJS last updated on 16/Feb/20

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

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