$ \int \sqrt{\frac{1-x}{1+x}} \mathrm{dx}= $

Question

$ \int \sqrt{\frac{1-x}{1+x}} \mathrm{dx}= $, $ \sin ^{-1} \mathrm{x}-\frac{1}{2} \sqrt{1-\mathrm{x}^{2}}+\mathrm{c} $, $ \sin ^{-1} \mathrm{x}+\frac{1}{2} \sqrt{1-\mathrm{x}^{2}}+\mathrm{c} $, $ \sin ^{-1} x-\sqrt{1-x^{2}}+c $, $ \sin ^{-1} \mathrm{x}+\sqrt{1-\mathrm{x}^{2}}+\mathrm{c} $

MARKS App · Accepted Answer

∫ 1 - x 1 + x d x Multiply and divide by 1 - x = ∫ 1 - x 1 - x 2 d x = ∫ d x 1 - x 2 - ∫ x 1 - x 2 d x = sin - 1 ⁡ x + 1 2 ∫ - 2 x 1 - x 2 d x = sin - 1 ⁡ x + 1 - x 2 + c

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