Answer
$=1-2\ln 2\approx-0.386$
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Work Step by Step
Find the indefinite integral first,
$\displaystyle \frac{x-1}{x+1}=\frac{x+1-2}{x+1}=\frac{x+1}{x+1}-\frac{2}{x+1}\\=1-\dfrac{2}{x+1}$
$\displaystyle \int\frac{x-1}{x+1}dx=\int dx-2\int\frac{1}{x+1}dx$
$=x-2\ln|x+1|+C$
Now, the definite integral:
$\displaystyle \int_{0}^{1}\frac{x-1}{x+1}dx=$
$=[x-2\ln|x+1|]_{0}^{1}$
$=1-2\ln 2-(0-2\ln 1)$
$=1-2\ln 2\approx-0.386$
