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Evaluate `int_0^1 (1-x)/(1+x) dx`A. `(1)/(2)log2`B. `(2log2+1)`C. `(2log2-1)`D. `((1)/(2)log2-1)` |
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Answer» On dividing `(-x+1)` by `(x+1)` we get : `I=int_(0)^(1){-1+(2)/((x+1))}dx=[-x+2log(x+1)]_(0)^(1)=(2log2-1)`. |
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