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Question Number 167360 by LEKOUMA last updated on 14/Mar/22

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

Answered by MJS_new last updated on 14/Mar/22

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

Commented by peter frank last updated on 14/Mar/22

$$\mathrm{thank}\mathrm{you}$$

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