The function $$F(x)=$$∫$$0x$$ log⁡$$1$$−$$t1+tdt$$ is

Correct option is 1an even function

Questions

The function $F (x) = \int_{0}^{x} \log \frac{1 - t}{1 + t} d t$ is

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an even function

an odd function

a periodic function

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detailed solution

Correct option is A

F(−x)=∫0−x log⁡1−t1+tdt=−∫0x log⁡1+u1−udu(u=−t)=∫0x log⁡1−u1+udu=F(x)

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The function F(x)=∫0x log⁡1−t1+tdt is

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The function $F (x) = \int_{0}^{x} \log \frac{1 - t}{1 + t} d t$ is