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Exercise 11.13 · Q16

Q.∫1−x1+x dx\displaystyle\int \sqrt{\dfrac{1-x}{1+x}}\,dx is

(1) 1−x2+sin⁡−1x+c\sqrt{1-x^2}+\sin^{-1}x+c
(2) sin⁡−1x−1−x2+c\sin^{-1}x-\sqrt{1-x^2}+c
(3) log⁡∣x+1−x2∣−1−x2+c\log\left|x+\sqrt{1-x^2}\right|-\sqrt{1-x^2}+c
(4) 1−x2+log⁡∣x+1−x2∣+c\sqrt{1-x^2}+\log\left|x+\sqrt{1-x^2}\right|+c
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Rationalise the surd, then split into two standard pieces.

Step 1. Multiply inside the root by 1−x1−x\dfrac{1-x}{1-x}: 1−x1+x=(1−x)21−x2=1−x1−x2\sqrt{\dfrac{1-x}{1+x}}=\sqrt{\dfrac{(1-x)^2}{1-x^2}}=\dfrac{1-x}{\sqrt{1-x^2}} for −1<x<1-1<x<1.

Step 2. Split: ∫1−x1−x2 dx=∫dx1−x2−∫x dx1−x2\displaystyle\int\dfrac{1-x}{\sqrt{1-x^2}}\,dx=\int\dfrac{dx}{\sqrt{1-x^2}}-\int\dfrac{x\,dx}{\sqrt{1-x^2}}. …

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