If int x(1+x) log(1+x^2) dx = F(x) log(1+x^2) | vedclass

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(C) Let $I = \int (x+x^2) \log(1+x^2) dx$. Using integration by parts, taking $\log(1+x^2)$ as the first function:$I = \log(1+x^2) \int (x+x^2) dx - \

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$\frac{x^2}{2} + \frac{x^3}{3} + \frac{1}{2}$

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(C) Let $I = \int (x+x^2) \log(1+x^2) dx$. Using integration by parts, taking $\log(1+x^2)$ as the first function:$I = \log(1+x^2) \int (x+x^2) dx - \int \left( \frac{d}{dx} \log(1+x^2) \int (x+x^2) dx ight) dx$$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} ight) - \int \frac{2x}{1+x^2} \left( \frac{x^2}{2} + \frac{x^3}{3} ight) dx$$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} ight) - \int \left( \frac{x^3}{1+x^2} + \frac{2x^4}{3(1+x^2)} ight) dx$Evaluating the integrals:$\int \frac{x^3}{1+x^2} dx = \int \frac{x(x^2+1-1)}{1+x^2} dx = \int x dx - \int \frac{x}{1+x^2} dx = \frac{x^2}{2} - \frac{1}{2} \log(1+x^2)$$\int \frac{2x^4}{3(1+x^2)} dx = \frac{2}{3} \int \frac{x^4-1+1}{1+x^2} dx = \frac{2}{3} \int (x^2-1) dx + \frac{2}{3} \int \frac{1}{1+x^2} dx = \frac{2x^3}{9} - \frac{2x}{3} + \frac{2}{3} 	an^{-1} x$Substituting back:$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} ight) - \left( \frac{x^2}{2} - \frac{1}{2} \log(1+x^2) + \frac{2x^3}{9} - \frac{2x}{3} + \frac{2}{3} 	an^{-1} x ight) + c$$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} + \frac{1}{2} ight) - \frac{2}{3} 	an^{-1} x - \frac{2x^3}{9} - \frac{x^2}{2} + \frac{2x}{3} + c$Comparing with the given expression, $F(x) = \frac{x^2}{2} + \frac{x^3}{3} + \frac{1}{2}$.

If $\int x(1+x) \log(1+x^2) dx = F(x) \log(1+x^2) - \frac{2}{3} \tan^{-1} x - \frac{2x^3}{9} - \frac{x^2}{2} + \frac{2x}{3} + c$, then $F(x) =$

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