If f(x) = int frac{x^2 + sin^2 x}{1 + x^2} cd | vedclass

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(D) Given $f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \cdot \sec^2 x \, dx$. We can rewrite the numerator as $x^2 + 1 - 1 + \sin^2 x = (1 + x^2) - (1

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

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(D) Given $f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \cdot \sec^2 x \, dx$. We can rewrite the numerator as $x^2 + 1 - 1 + \sin^2 x = (1 + x^2) - (1 - \sin^2 x) = (1 + x^2) - \cos^2 x$. Substituting this into the integral: $f(x) = \int \frac{(1 + x^2) - \cos^2 x}{1 + x^2} \cdot \sec^2 x \, dx$ $f(x) = \int \left( 1 - \frac{\cos^2 x}{1 + x^2} ight) \sec^2 x \, dx$ $f(x) = \int \sec^2 x \, dx - \int \frac{\cos^2 x \cdot \sec^2 x}{1 + x^2} \, dx$ Since $\cos^2 x \cdot \sec^2 x = 1$,we have: $f(x) = \int \sec^2 x \, dx - \int \frac{1}{1 + x^2} \, dx$ $f(x) = 	an x - 	an^{-1} x + C$. Given $f(0) = 0$: $0 = 	an(0) - 	an^{-1}(0) + C \implies 0 = 0 - 0 + C \implies C = 0$. Thus,$f(x) = 	an x - 	an^{-1} x$. Therefore,$f(1) = 	an(1) - 	an^{-1}(1) = 	an(1) - \frac{\pi}{4}$.

If $f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \cdot \sec^2 x \, dx$ and $f(0) = 0$,then $f(1) = $

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