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 $I = \int \frac{3-x^2}{1-2x+x^2} e^x dx$.We can rewrite the denominator as $(1-x)^2$.So,$I = \int \frac{3-x^2}{(1-x)^2} e^x dx$

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

Vedclass · Answer

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

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