int e^x left( frac{1-x}{1+x^2} right)^2 dx = | vedclass

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(B) The given integral is $I = \int e^x \left( \frac{1-x}{1+x^2} \right)^2 dx$.Expanding the numerator,we get $I = \int e^x \frac{1-2x+x^2}{(1+x^2)^2}

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$\frac{e^x}{1+x^2}$

Vedclass · Answer

(B) The given integral is $I = \int e^x \left( \frac{1-x}{1+x^2} \right)^2 dx$.Expanding the numerator,we get $I = \int e^x \frac{1-2x+x^2}{(1+x^2)^2} dx$.This can be rewritten as $I = \int e^x \left[ \frac{1+x^2}{(1+x^2)^2} - \frac{2x}{(1+x^2)^2} \right] dx$.$I = \int e^x \left[ \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} \right] dx$.We know the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.Here,let $f(x) = \frac{1}{1+x^2}$.Then,$f'(x) = \frac{d}{dx} (1+x^2)^{-1} = -1(1+x^2)^{-2} \cdot (2x) = -\frac{2x}{(1+x^2)^2}$.Since the integrand is in the form $e^x [f(x) + f'(x)]$,the solution is $e^x f(x) + C = \frac{e^x}{1+x^2} + C$.

$\int e^x \left( \frac{1-x}{1+x^2} \right)^2 dx = $ . . . . . . + $C$

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