Let f(x) = int frac{x^2 dx}{(1 + x^2)(1 + sqr | vedclass

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(B) Given $f(x) = \int \frac{x^2 dx}{(1 + x^2)(1 + \sqrt{1 + x^2})}$.Let $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.Substituting these int

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$\log(1 + \sqrt{2}) - \frac{\pi}{4}$

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(B) Given $f(x) = \int \frac{x^2 dx}{(1 + x^2)(1 + \sqrt{1 + x^2})}$.Let $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.Substituting these into the integral:$f(x) = \int \frac{	an^2 	heta \cdot \sec^2 	heta \, d	heta}{\sec^2 	heta (1 + \sec 	heta)} = \int \frac{	an^2 	heta \, d	heta}{1 + \sec 	heta}$.Using $	an^2 	heta = \sec^2 	heta - 1$,we get:$f(x) = \int \frac{\sec^2 	heta - 1}{1 + \sec 	heta} \, d	heta = \int \frac{(\sec 	heta - 1)(\sec 	heta + 1)}{1 + \sec 	heta} \, d	heta = \int (\sec 	heta - 1) \, d	heta$.Integrating,we get $f(x) = \log|\sec 	heta + 	an 	heta| - 	heta + C$.Since $x = 	an 	heta$,then $\sec 	heta = \sqrt{1 + x^2}$ and $	heta = 	an^{-1} x$.Thus,$f(x) = \log|\sqrt{1 + x^2} + x| - 	an^{-1} x + C$.Given $f(0) = 0$,we have $\log(1) - 0 + C = 0$,which implies $C = 0$.Therefore,$f(x) = \log(x + \sqrt{1 + x^2}) - 	an^{-1} x$.For $x = 1$,$f(1) = \log(1 + \sqrt{2}) - 	an^{-1}(1) = \log(1 + \sqrt{2}) - \frac{\pi}{4}$.

Let $f(x) = \int \frac{x^2 dx}{(1 + x^2)(1 + \sqrt{1 + x^2})}$ and $f(0) = 0$,then the value of $f(1)$ is:

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