Mathematics

$$\displaystyle \underset{1}{\overset{2}{\int}} \dfrac{1 - x}{1 + x} dx$$ equals 


ANSWER

$$2 \, \log \, \left(\dfrac{3}{2} \right) - 1$$


SOLUTION
$$\int \dfrac{1-x}{1+x}dx$$

$$u=1+x\rightarrow du=dx$$

$$=\int \dfrac{-u+2}{u}du$$

$$=\int -1+\dfrac{2}{u}du$$

$$=-u+2\log |u| +C$$

$$=-(1+x)+2\log |(1+x)| +C$$

$$\int_{1}^{2}\dfrac{1-x}{1+x}dx=(2\log 3 -3) -(2\log 2-2)$$

$$=\log \left ( \dfrac{9}{4} \right )-1$$

$$=\log \left ( \dfrac{3}{2} \right )^2-1$$

$$=\left ( 2\right ) \log \left ( \dfrac{3}{2} \right )-1$$
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Single Correct Medium Published on 17th 09, 2020
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