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Question Number 91086 by jagoll last updated on 28/Apr/20

$$\int \frac{2-x^2}{1+x\sqrt{1-x^2}}dx$$

Answered by MJS last updated on 28/Apr/20

$$\int \frac{2-x^2}{1+x\sqrt{1-x^2}}dx=$$$$\mathrm{the}\mathrm{two}\mathrm{necessary}\mathrm{steps}\mathrm{in}\mathrm{one}:$$$$x=\sin \theta \wedge \theta =2\arctan t\Rightarrow x=\frac{2t}{t^2+1}$$$$t=\frac{1+\sqrt{1-x^2}}{x}\to dx=-\frac{x^2\sqrt{1-x^2}}{1+\sqrt{1-x^2}}dt=-\frac{2(t^2-1)}{{(t^2+1)}^2}dt$$$$=-4\int \frac{(t^2-1)(t^4+1)}{{(t^2+1)}^2(t^4+2t^3+2t^2-2t+1)}dt=$$$$t^4+2t^3+2t^2-2t+1=(t^2+(1-\sqrt{3})t+2-\sqrt{3})(t^2+(1+\sqrt{3})t+2+\sqrt{3})$$$$=4\int \frac{t}{{(t^2+1)}^2}dt-\int \frac{\sqrt{3}t-1}{t^2+(1-\sqrt{3})t+2-\sqrt{3}}dt+\int \frac{\sqrt{3}t+1}{t^2+(1+\sqrt{3})t+2+\sqrt{3}}dt=$$$$=-\frac{2}{t^2+1}+\frac{\sqrt{3}}{2}\ln \frac{t^2+(1+\sqrt{3})t+2+\sqrt{3}}{t^2+(1-\sqrt{3})t+2-\sqrt{3}}+\arctan ((1-\sqrt{3})t-1)+\arctan ((1+\sqrt{3})t+1)=$$$$=-\frac{x^2}{1+\sqrt{1-x^2}}+\frac{\sqrt{3}}{2}\ln \frac{x^2-2+\sqrt{3(1-x^2)}}{x^2-\sqrt{3}x+1}+\arctan \frac{(1-\sqrt{3})(1+\sqrt{1-x^2})-x}{x}+\arctan \frac{(1+\sqrt{3})(1+\sqrt{1-x^2})-x}{x}+C$$$$\mathrm{I}\mathrm{found}\mathrm{no}\mathrm{easier}\mathrm{path}$$$$\mathrm{I}\mathrm{hope}\mathrm{I}\mathrm{made}\mathrm{no}\mathrm{typos}...$$

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