int_{0}^{1} frac{x^{2}}{1+x^{2}} , dx = | vedclass

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(B) Let $I = \int_{0}^{1} \frac{x^{2}}{1+x^{2}} \, dx$. We can rewrite the integrand as $\frac{x^{2}+1-1}{1+x^{2}} = \frac{1+x^{2}}{1+x^{2}} - \frac{1

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int_{0}^{1} \frac{x^{2}}{1+x^{2}} \, dx$. We can rewrite the integrand as $\frac{x^{2}+1-1}{1+x^{2}} = \frac{1+x^{2}}{1+x^{2}} - \frac{1}{1+x^{2}} = 1 - \frac{1}{1+x^{2}}$. Now,integrate term by term: $I = \int_{0}^{1} \left( 1 - \frac{1}{1+x^{2}} ight) \, dx$. $I = \left[ x - 	an^{-1}(x) ight]_{0}^{1}$. Evaluating at the limits: $I = (1 - 	an^{-1}(1)) - (0 - 	an^{-1}(0))$. Since $	an^{-1}(1) = \frac{\pi}{4}$ and $	an^{-1}(0) = 0$,$I = 1 - \frac{\pi}{4} - 0 = 1 - \frac{\pi}{4}$.

$\int_{0}^{1} \frac{x^{2}}{1+x^{2}} \, dx =$

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