Question

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

• A. $\log 4$
• B. $\log \left(\frac{4}{e}\right)$
• C. $1$
• D. $\log \left(\frac{e}{4}\right)$

Answer 1

The correct option is B $\log \left(\frac{4}{e}\right)$

${\int }_0^1\left(\frac{1-x}{1+x}\right)dx$

$\int \frac{1-x}{1+x}dx=\int \frac{2}{1+x}dx-\int 1\cdot dx$

$=2\log (1+x)-x+c$

$\int \frac{1-x}{1+x}dx=2\log (1+x)-x+c$

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

$=2\log \left(2\right)-1$

$=\log 4-\log e$

$=\log \left(\frac{4}{e}\right)$