Question

${\int }_1^2\frac{1-x}{1+x}dx$.

Answer 1

We are given that,

$I={\int }_1^2\frac{1-x}{1+x}dx$

$={\int }_1^2\frac{1-(x+1)+1}{(1+x)}dx$

$={\int }_1^2\frac{2-(1+x)}{(1+x)}dx$

$={\int }_1^2\frac{2}{(1+x)}dx-{\int }_1^{2}1.dx$

$=[2\ln |1+x|{]}_1^2-[x{]}_1^2$

$=2\ln \left(3\right)-2\ln \left(2\right)-(2-1)$

$=2\left(\ln \right(3)-\ln (2\left)\right)-1$

$I=2\ln \left(\frac{3}{2}\right)-1$