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Question
Evaluate ∫1−1log2−x2+xdx.
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Solution
Let I=∫1−1log2−x2+xdx
And f(x)=loge2−x2+x
Then f(−x)=log2+x2−x=log(2−x2+x)−1=−log2−x2+x=−f(x)
Hence I=∫1−1loge(2−x2+x)dx=0
And f(x)=loge2−x2+x
Then f(−x)=log2+x2−x=log(2−x2+x)−1=−log2−x2+x=−f(x)
Hence I=∫1−1loge(2−x2+x)dx=0
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