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$\int_{0}^{1}\left(\frac{x^{2}-2}{x^{2}+1}\right) d x=$
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The correct answer is:
$1-\frac{3 \pi}{4}$
(C)
$\begin{aligned} \int_{0}^{1} \frac{x^{2}+1-3}{x^{2}+1} d x &=\int_{0}^{1} \frac{x^{2}+1}{x^{2}+1}-\frac{3}{x^{2}+1} d x \\ &=\int_{0}^{1} 1-\frac{3}{x^{2}+1} d x=\left[x-3 \tan ^{-1} x\right]_{0}^{1} \\ &=\left(1-3 \tan ^{-1} 1\right)-\left(0-3 \tan ^{-1} 0\right)=1-3 \frac{\pi}{4} \end{aligned}$
$\begin{aligned} \int_{0}^{1} \frac{x^{2}+1-3}{x^{2}+1} d x &=\int_{0}^{1} \frac{x^{2}+1}{x^{2}+1}-\frac{3}{x^{2}+1} d x \\ &=\int_{0}^{1} 1-\frac{3}{x^{2}+1} d x=\left[x-3 \tan ^{-1} x\right]_{0}^{1} \\ &=\left(1-3 \tan ^{-1} 1\right)-\left(0-3 \tan ^{-1} 0\right)=1-3 \frac{\pi}{4} \end{aligned}$
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