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

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(A) Given $f(x) = \int \left( \frac{x^2 + \sin^2 x}{1 + x^2} \right) \sec^2 x \, dx$.We can rewrite the integrand as:$f(x) = \int \frac{x^2 \sec^2 x +

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

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(A) Given $f(x) = \int \left( \frac{x^2 + \sin^2 x}{1 + x^2} ight) \sec^2 x \, dx$.We can rewrite the integrand as:$f(x) = \int \frac{x^2 \sec^2 x + \sin^2 x \sec^2 x}{1 + x^2} \, dx$Since $\sin^2 x \sec^2 x = 	an^2 x$,we have:$f(x) = \int \frac{x^2 \sec^2 x + 	an^2 x}{1 + x^2} \, dx$Using $\sec^2 x = 1 + 	an^2 x$,we get:$f(x) = \int \frac{x^2(1 + 	an^2 x) + 	an^2 x}{1 + x^2} \, dx$$f(x) = \int \frac{x^2 + x^2 	an^2 x + 	an^2 x}{1 + x^2} \, dx = \int \frac{x^2 + 	an^2 x(1 + x^2)}{1 + x^2} \, dx$$f(x) = \int \frac{x^2}{1 + x^2} \, dx + \int 	an^2 x \, dx$$f(x) = \int \frac{x^2 + 1 - 1}{1 + x^2} \, dx + \int (\sec^2 x - 1) \, dx$$f(x) = \int (1 - \frac{1}{1 + x^2}) \, dx + \int \sec^2 x \, dx - \int 1 \, dx$$f(x) = x - 	an^{-1} x + 	an x - x + C = 	an x - 	an^{-1} x + C$Given $f(0) = 0$,we have $0 = 	an 0 - 	an^{-1} 0 + C$,so $C = 0$.Thus,$f(x) = 	an x - 	an^{-1} x$.Therefore,$f(1) = 	an 1 - 	an^{-1}(1) = 	an 1 - \frac{\pi}{4}$.

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

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