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

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Question

(A) $	ext{આપણે જાણીએ છીએ કે}$ $\int e^{x}[f(x) + f'(x)] dx = e^{x}f(x) + C$.$	ext{આપેલ સંકલન}$ $I = \int e^{x} \left( \frac{1-x}{1+x^{2}} ight)^{2

Vedclass · Accepted Answer

$e^{x}\left(\frac{1}{1+x^{2}}\right)+C$

Vedclass · Answer

(A) $	ext{આપણે જાણીએ છીએ કે}$ $\int e^{x}[f(x) + f'(x)] dx = e^{x}f(x) + C$.$	ext{આપેલ સંકલન}$ $I = \int e^{x} \left( \frac{1-x}{1+x^{2}} ight)^{2} dx$ $	ext{છે।}$$	ext{અંશનું વિસ્તરણ કરતા:}$ $I = \int e^{x} \frac{1 + x^{2} - 2x}{(1+x^{2})^{2}} dx$.$	ext{અપૂર્ણાંકને અલગ કરતા:}$ $I = \int e^{x} \left[ \frac{1+x^{2}}{(1+x^{2})^{2}} - \frac{2x}{(1+x^{2})^{2}} ight] dx$.$I = \int e^{x} \left[ \frac{1}{1+x^{2}} - \frac{2x}{(1+x^{2})^{2}} ight] dx$.$	ext{ધારો કે}$ $f(x) = \frac{1}{1+x^{2}}$. $	ext{તો}$ $f'(x) = -\frac{1}{(1+x^{2})^{2}} \cdot (2x) = -\frac{2x}{(1+x^{2})^{2}}$.$	ext{આમ,}$ $I = \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.$	ext{તેથી,}$ $I = \frac{e^{x}}{1+x^{2}} + C$.

$\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} \,d x=$

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