The value of the definite integral $\int_0^1 e^x \frac{(1-x)^2}{(1+x^2)^2}dx$ is:

The definite integral to evaluate is $\int_0^1 e^x \frac{(1-x)^2}{(1+x^2)^2}dx$.

Expand $(1-x)^2$:

$(1-x)^2 = 1 - 2x + x^2$

The integral becomes:

$\int_0^1 e^x \frac{1 - 2x + x^2}{(1+x^2)^2}dx$

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Regroup the numerator as $(1 + x^2) - 2x$:

$\int_0^1 e^x \frac{1 + x^2 - 2x}{(1+x^2)^2}dx$

Split into two integrals:

$\int_0^1 e^x \frac{1 + x^2}{(1+x^2)^2}dx - \int_0^1 e^x \frac{2x}{(1+x^2)^2}dx$

$= \int_0^1 \frac{e^x}{1+x^2}dx - \int_0^1 e^x \frac{2x}{(1+x^2)^2}dx$

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Note that:

$\frac{d}{dx}\left[\frac{1}{1+x^2}\right] = \frac{-2x}{(1+x^2)^2}$

Therefore:

$\frac{2x}{(1+x^2)^2} = -\frac{d}{dx}\left[\frac{1}{1+x^2}\right]$

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For the second integral, apply integration by parts:

$\int_0^1 e^x \frac{d}{dx}\left[\frac{1}{1+x^2}\right]dx = \left[e^x \cdot \frac{1}{1+x^2}\right]_0^1 - \int_0^1 e^x \frac{1}{1+x^2}dx$

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Substituting back:

$\int_0^1 \frac{e^x}{1+x^2}dx - \left(\left[e^x \cdot \frac{1}{1+x^2}\right]_0^1 - \int_0^1 \frac{e^x}{1+x^2}dx\right)$

The two $\int \frac{e^x}{1+x^2}dx$ terms cancel:

$= -\left[e^x \cdot \frac{1}{1+x^2}\right]_0^1$

$= \left[e^x \cdot \frac{1}{1+x^2}\right]_0^1$

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Evaluate at the limits:

At $x = 1$: $\frac{e^1}{1+1^2}$

$= \frac{e}{2}$

At $x = 0$: $\frac{e^0}{1+0^2}$

$= \frac{1}{1}$

$= 1$

Therefore:

$\frac{e}{2} - 1$