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Problem The value of ∫ log ( 1 − x + 1 + x ) d x \int \log(\sqrt{1-x} + \sqrt{1+x}) dx ∫ log ( 1 − x + 1 + x ) d x , is equal to
a. x log ( 1 − x + 1 + x ) + 1 2 x − 1 2 sin − 1 ( x ) + C x \log(\sqrt{1-x} + \sqrt{1+x}) + \frac{1}{2}x - \frac{1}{2}\sin^{-1}(x) + C x log ( 1 − x + 1 + x ) + 2 1 x − 2 1 sin − 1 ( x ) + C
b. x log ( 1 − x + 1 + x ) + 1 2 x + 1 2 sin − 1 ( x ) + C x \log(\sqrt{1-x} + \sqrt{1+x}) + \frac{1}{2}x + \frac{1}{2}\sin^{-1}(x) + C x log ( 1 − x + 1 + x ) + 2 1 x + 2 1 sin − 1 ( x ) + C
c. x log ( 1 − x + 1 + x ) − 1 2 x + 1 2 sin − 1 ( x ) + C x \log(\sqrt{1-x} + \sqrt{1+x}) - \frac{1}{2}x + \frac{1}{2}\sin^{-1}(x) + C x log ( 1 − x + 1 + x ) − 2 1 x + 2 1 sin − 1 ( x ) + C
d. None of the above
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