Answer
$-\ln 3\approx-1.099$
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Work Step by Step
Find the indefinite integral first,
$x^{2}-2=(x^{2}-1)-1,$
$\displaystyle \frac{x^{2}-2}{x+1}=\frac{(x^{2}-1)-1}{x+1}=\frac{(x^{2}-1)}{x+1}-\frac{1}{x+1}=\frac{(x+1)(x-1)}{x+1}-\frac{1}{x+1}=$
$=x-1-\displaystyle \frac{1}{x+1}, \ \ \ $so
$\displaystyle \int\frac{x^{2}-2}{x+1}dx=\int(x-1-\frac{1}{x+1})dx$
$=\displaystyle \frac{1}{2}x^{2}-x-\ln|x+1|+C$
Now, the definite integral:
$\displaystyle \int_{0}^{2}\frac{x^{2}-2}{x+1}dx=\left[\frac{1}{2}x^{2}-x-\ln|x+1|\right]_{0}^{2}$
$=\displaystyle \frac{1}{2}\cdot 4-2-\ln 3-(0-0-\ln 1)$
$=-\ln 3\approx-1.099$
