A. $$ \int \frac{(2 x-1) e^{x^{2}}}{e^{x}} d x $$

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The integral simplifies to $$ e^{x^2 - x} + C $$. To solve the integral $$ \int \frac{(2x-1)e^{x^2}}{e^x} \, dx, $$ we can simplify the integrand first. We rewrite the expression: $$ \frac{(2x-1)e^{x^2}}{e^x} = (2x-1)e^{x^2 - x}. $$ Thus, the integral becomes: $$ \int (2x-1)e^{x^2 - x} \, dx. $$ Next, we can use substitution to simplify the integral. Let: $$ u = x^2 - x. $$ Then, we compute the derivative $$du$$: $$ du = (2x - 1) \, dx. $$ This means that $$dx = \frac{du}{2x - 1}$$. Now we can rewrite the integral in terms of $$u$$: $$ \int e^u \, du. $$ The integral of $$e^u$$ is simply: $$ e^u + C, $$ where $$C$$ is the constant of integration. Now we substitute back for $$u$$: $$ e^{x^2 - x} + C. $$ Thus, the final result for the integral is: $$ \int \frac{(2x-1)e^{x^2}}{e^x} \, dx = e^{x^2 - x} + C. $$

A. \( \int \frac{(2 x-1) e^{x^{2}}}{e^{x}} d x \)

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A. \( \int \frac{(2 x-1) e^{x^{2}}}{e^{x}} d x \)

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