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$\int_0^1 \frac{x e^x}{(x+1)^2} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{e}{2}-1$
$I=\int_0^1 \frac{x e^x}{(x+1)^2} d x$
$$
\begin{aligned}
I & =\int_0^1 e^x\left[\frac{x}{(x+1)^2}\right] d x \\
& =\int_0^1 e^x\left[\frac{1}{x+1}-\frac{1}{(x+1)^2}\right] d x
\end{aligned}
$$
Let $f(x)=\frac{1}{x+1} \Rightarrow f^{\prime}(x)=\frac{-1}{(x+1)^2}$
$\begin{aligned} \therefore \quad I & =\int_0^1 e^x\left[f(x)+f^{\prime}(x)\right] d x \\ \Rightarrow \quad I & =\left.e^x f(x)\right|_0 ^1 \\ & I=\left[\frac{e^x}{x+1}\right]_0^1=\frac{e}{2}-1\end{aligned}$
$$
\begin{aligned}
I & =\int_0^1 e^x\left[\frac{x}{(x+1)^2}\right] d x \\
& =\int_0^1 e^x\left[\frac{1}{x+1}-\frac{1}{(x+1)^2}\right] d x
\end{aligned}
$$
Let $f(x)=\frac{1}{x+1} \Rightarrow f^{\prime}(x)=\frac{-1}{(x+1)^2}$
$\begin{aligned} \therefore \quad I & =\int_0^1 e^x\left[f(x)+f^{\prime}(x)\right] d x \\ \Rightarrow \quad I & =\left.e^x f(x)\right|_0 ^1 \\ & I=\left[\frac{e^x}{x+1}\right]_0^1=\frac{e}{2}-1\end{aligned}$
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