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Question Number 88559 by 675480065 last updated on 11/Apr/20

$${\int }_{\frac{-\pi }{6}}^{\frac{3\pi }{4}}\frac{\sqrt{tanx}}{1+\sqrt{tanx}}dx$$

Commented by MJS last updated on 11/Apr/20

$$\mathrm{t}=\sqrt{\tan x}\to dx={2\cos }^{2}x\sqrt{\tan x}dt$$$$\mathrm{this}\mathrm{leads}\mathrm{to}$$$$2\int \frac{t^2}{(t^4+1)(t+1)}dt=$$$$=\int \frac{dt}{t+1}-\frac{1-\sqrt{2}}{2}\int \frac{t+1}{t^2-\sqrt{2}t+1}dt-\frac{1+\sqrt{2}}{2}\int \frac{t+1}{t^2+\sqrt{2}t+1}dt$$$$\mathrm{now}\mathrm{use}\mathrm{formulas}$$$$\mathrm{but}\mathrm{with}-\frac{\pi }{6}⩽x⩽\frac{3\pi }{4}\mathrm{the}\mathrm{integral}\notin \mathbb{R}$$

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