If int frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x | vedclass

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(A) Given the integral $I = \int \frac{3-x^2}{1-2 x+x^2} e^x d x$.Since $1-2x+x^2 = (1-x)^2$,we can rewrite the integrand as:$I = \int \frac{3-x^2}{(1

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$\frac{1+x}{1-x}$

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(A) Given the integral $I = \int \frac{3-x^2}{1-2 x+x^2} e^x d x$.Since $1-2x+x^2 = (1-x)^2$,we can rewrite the integrand as:$I = \int \frac{3-x^2}{(1-x)^2} e^x d x$We can manipulate the numerator to match the form $f(x) + f'(x)$:$I = \int \frac{2 + 1 - x^2}{(1-x)^2} e^x d x = \int \left( \frac{2}{(1-x)^2} + \frac{1-x^2}{(1-x)^2} \right) e^x d x$$I = \int \left( \frac{2}{(1-x)^2} + \frac{(1-x)(1+x)}{(1-x)^2} \right) e^x d x$$I = \int \left( \frac{2}{(1-x)^2} + \frac{1+x}{1-x} \right) e^x d x$Let $f(x) = \frac{1+x}{1-x}$. Then $f'(x) = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{1-x+1+x}{(1-x)^2} = \frac{2}{(1-x)^2}$.Since $\int (f(x) + f'(x)) e^x d x = e^x f(x) + c$,we have:$I = e^x \left( \frac{1+x}{1-x} \right) + c$.Comparing this with the given expression $e^x f(x) + c$,we get $f(x) = \frac{1+x}{1-x}$.

If $\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c$,then find $f(x)$.

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