Given that frac{d}{d x}left(tan ^{-1} xright) | vedclass

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(C) We need to evaluate the integral $I = \int \frac{3 x^6-2 x^4+x^2-2}{x^2+1} d x$. Performing polynomial long division of the numerator $3x^6 - 2x^4

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$\frac{3}{5} x^5-\frac{5}{3} x^3+6 x-8 	an ^{-1} x+c$

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(C) We need to evaluate the integral $I = \int \frac{3 x^6-2 x^4+x^2-2}{x^2+1} d x$. Performing polynomial long division of the numerator $3x^6 - 2x^4 + x^2 - 2$ by the denominator $x^2 + 1$: $3x^6 - 2x^4 + x^2 - 2 = (x^2 + 1)(3x^4 - 5x^2 + 6) - 8$. Thus, the integral becomes: $I = \int \left( 3x^4 - 5x^2 + 6 - \frac{8}{x^2+1} ight) d x$. Integrating term by term: $I = 3 \int x^4 d x - 5 \int x^2 d x + 6 \int 1 d x - 8 \int \frac{1}{x^2+1} d x$. Using the standard integral formulas: $I = 3 \left( \frac{x^5}{5} ight) - 5 \left( \frac{x^3}{3} ight) + 6x - 8 	an^{-1} x + c$. $I = \frac{3}{5} x^5 - \frac{5}{3} x^3 + 6x - 8 	an^{-1} x + c$.

Given that $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$ and $\frac{d}{d x}\left(\sin h^{-1} x\right)=\frac{1}{\sqrt{1+x^2}}$. Then $\int \frac{3 x^6-2 x^4+x^2-2}{x^2+1} d x=$

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