If x 0 and x neq (2n+1) frac{pi}{2}, then i | vedclass

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(D) Let $I = \int \left(x \sqrt{x} - e^{\log(\sec x 	an x)} + \frac{3x^2 - 2x + 1}{x^2}ight) dx$. Using the property $e^{\log f(x)} = f(x)$, we get

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$\frac{2}{5} x^2 \sqrt{x} - \sec x + 3x - 2 \log x - \frac{1}{x} + c$

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(D) Let $I = \int \left(x \sqrt{x} - e^{\log(\sec x 	an x)} + \frac{3x^2 - 2x + 1}{x^2}ight) dx$. Using the property $e^{\log f(x)} = f(x)$, we get: $I = \int \left(x^{3/2} - \sec x 	an x + 3 - \frac{2}{x} + \frac{1}{x^2}ight) dx$. Integrating term by term: $\int x^{3/2} dx = \frac{x^{5/2}}{5/2} = \frac{2}{5} x^2 \sqrt{x}$. $\int -\sec x 	an x dx = -\sec x$. $\int 3 dx = 3x$. $\int -\frac{2}{x} dx = -2 \log x$. $\int \frac{1}{x^2} dx = \int x^{-2} dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$. Combining these, we get $I = \frac{2}{5} x^2 \sqrt{x} - \sec x + 3x - 2 \log x - \frac{1}{x} + c$.

If $x > 0$ and $x \neq (2n+1) \frac{\pi}{2}$, then $\int \left(x \sqrt{x} - e^{\log(\sec x \tan x)} + \frac{3x^2 - 2x + 1}{x^2}\right) dx =$

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