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The value of the integral $\int \mathrm{e}^{\mathrm{x}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}^{2}}\right)^{2} \mathrm{~d} \mathrm{x}$ is
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The correct answer is:
$\frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{x}^{2}}+\mathrm{C}$
Let $I=\int e^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x$
$\Rightarrow I=\int e^{x} \frac{1+x^{2}-2 x}{\left(1+x^{2}\right)^{2}} d x$
$\quad=\int e^{x}\left[\frac{1+x^{2}}{\left(1+x^{2}\right)^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x$
$\Rightarrow I=\int e^{x} \frac{1}{\left(1+x^{2}\right)} d x-2 \int e^{x} \cdot \frac{x}{\left(1+x^{2}\right)^{2}} d x$
$\Rightarrow I=\frac{1}{\left(1+x^{2}\right)} \cdot e^{x}-\int e^{x} \cdot\left[-\left(1+x^{2}\right)^{-2}\right] 2 x d x$
$\quad-2 \int \frac{e^{x} x}{\left(1+x^{2}\right)^{2}} d x$
$\Rightarrow I=\frac{e^{x}}{1+x^{2}}+2 \int \frac{e^{x} x}{\left(1+x^{2}\right)^{2}} d x -2\int\frac{\mathrm{e}^{\mathrm{x}} \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)^{2}} \mathrm{dx}$
$\Rightarrow I=\frac{e^{x}}{1+x^{2}}+C$
$\Rightarrow I=\int e^{x} \frac{1+x^{2}-2 x}{\left(1+x^{2}\right)^{2}} d x$
$\quad=\int e^{x}\left[\frac{1+x^{2}}{\left(1+x^{2}\right)^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x$
$\Rightarrow I=\int e^{x} \frac{1}{\left(1+x^{2}\right)} d x-2 \int e^{x} \cdot \frac{x}{\left(1+x^{2}\right)^{2}} d x$
$\Rightarrow I=\frac{1}{\left(1+x^{2}\right)} \cdot e^{x}-\int e^{x} \cdot\left[-\left(1+x^{2}\right)^{-2}\right] 2 x d x$
$\quad-2 \int \frac{e^{x} x}{\left(1+x^{2}\right)^{2}} d x$
$\Rightarrow I=\frac{e^{x}}{1+x^{2}}+2 \int \frac{e^{x} x}{\left(1+x^{2}\right)^{2}} d x -2\int\frac{\mathrm{e}^{\mathrm{x}} \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)^{2}} \mathrm{dx}$
$\Rightarrow I=\frac{e^{x}}{1+x^{2}}+C$
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