Lời giải

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

Xét \( I_1 = \int_0^1 \frac{1}{x+1} e^x \, dx \):  

Đặt \( 
\begin{cases}
u = \frac{1}{x+1} \\
dv = e^x \, dx 
\end{cases}
\Rightarrow
\begin{cases}
du = -\frac{1}{(x+1)^2} \, dx \\
v = e^x
\end{cases}\)

 \( I_1 = \int_0^1 \frac{1}{x+1} e^x \, dx = \frac{e^x}{x+1} \bigg|_0^1 + \int_0^1 \frac{1}{(x+1)^2} e^x \, dx \)

 \( \Rightarrow I = \frac{e^x}{x+1} \Bigg|_0^1 = \frac{e}{2} - 1 \)