int_{0}^{1} left(frac{x^{2}-2}{x^{2}+1}right) | vedclass

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(B) To evaluate the integral $I = \int_{0}^{1} \frac{x^{2}-2}{x^{2}+1} dx$,we can rewrite the numerator as $(x^{2}+1)-3$.$I = \int_{0}^{1} \frac{x^{2}

Vedclass · Accepted Answer

$1-\frac{3\pi}{4}$

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(B) To evaluate the integral $I = \int_{0}^{1} \frac{x^{2}-2}{x^{2}+1} dx$,we can rewrite the numerator as $(x^{2}+1)-3$.$I = \int_{0}^{1} \frac{x^{2}+1-3}{x^{2}+1} dx$$I = \int_{0}^{1} \left( \frac{x^{2}+1}{x^{2}+1} - \frac{3}{x^{2}+1} ight) dx$$I = \int_{0}^{1} \left( 1 - \frac{3}{x^{2}+1} ight) dx$Integrating term by term,we get:$I = [x - 3 	an^{-1}(x)]_{0}^{1}$Substituting the limits:$I = (1 - 3 	an^{-1}(1)) - (0 - 3 	an^{-1}(0))$Since $	an^{-1}(1) = \frac{\pi}{4}$ and $	an^{-1}(0) = 0$:$I = (1 - 3 \cdot \frac{\pi}{4}) - (0 - 0)$$I = 1 - \frac{3\pi}{4}$

$\int_{0}^{1} \left(\frac{x^{2}-2}{x^{2}+1}\right) dx =$

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