Question

Evaluate the following definite integral:

${\int }_0^1\frac{1-x}{1+x}dx$

• A. $2\log 2+1$
• B. $2\log 2$
• C. $2\log 2-1$
• D. $\log 4$

Answer 1

Answer: (C)

$2\log 2-1$

$I={\int }_0^1\frac{1-x}{1+x}dx$

$={\int }_0^1\frac{2-(1+x)}{1+x}dx$

$={\int }_0^1(\frac{2}{1+x}-1)dx$

$={\left[2\log \right(x+1)-x]}_0^1$

$\Rightarrow I=(2\log 2-1)-(2\log 1-0)$

$=2\log 2-1$