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Integrate from 0 to 1 (x/1+x)\(\int\limits_0^1\frac{x}{1+x}dx\) |
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Answer» \(\int\limits_0^1\frac{x}{1+x}dx\) = \(\int\limits_0^1\frac{1+x-1}{1+x}dx\) = \(\int\limits_0^1(1+\frac{1}{1+x})dx\) = \( [x - log(1 + x)]_0^1\) = (1 - 0) -(log 2 - log 1) = 1 - log 2 = log e - log 2 = log(e/2) |
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