If ∫_0^1(1-x)/(1+x) dx=Alog 2+B , then the value of A and B are

The correct answer is Option (3) → \( A=2,B=-1 \)

Since 1 − x = 2 − (1 + x), the numerator can be written as a multiple of the denominator plus a constant:

    \( \displaystyle \frac{1-x}{1+x}=\frac{2-(1+x)}{1+x} \)

    \( \displaystyle =\frac{2}{1+x}-1 \)

So

    \( \displaystyle \int_{0}^{1}\left(\frac{2}{1+x}-1\right)dx=\Big[2\log(1+x)-x\Big]_{0}^{1} \)

    \( \displaystyle =(2\log 2-1)-(2\log 1-0) \)

    \( \displaystyle =2\log 2-1 \)

Comparing with A log 2 + B gives A = 2 and B = −1.