2-x-2-1-x-1-x-2-dx-

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

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

Answered by MJS last updated on 28/Apr/20

\int \frac{2 - x^{2}}{1 + x \sqrt{1 - x^{2}}} d x =

the two necessary steps in one :

x = \sin θ \land θ = 2 \arctan t \Rightarrow x = \frac{2 t}{t^{2} + 1}

t = \frac{1 + \sqrt{1 - x^{2}}}{x} \to d x = - \frac{x^{2} \sqrt{1 - x^{2}}}{1 + \sqrt{1 - x^{2}}} d t = - \frac{2 (t^{2} - 1)}{{(t^{2} + 1)}^{2}} d t

= - 4 \int \frac{(t^{2} - 1) (t^{4} + 1)}{{(t^{2} + 1)}^{2} (t^{4} + 2 t^{3} + 2 t^{2} - 2 t + 1)} d t =

t^{4} + 2 t^{3} + 2 t^{2} - 2 t + 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}} d t - \int \frac{\sqrt{3} t - 1}{t^{2} + (1 - \sqrt{3}) t + 2 - \sqrt{3}} d t + \int \frac{\sqrt{3} t + 1}{t^{2} + (1 + \sqrt{3}) t + 2 + \sqrt{3}} d t =

= - \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

I found no easier path

I hope I made no typos \dots

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