Ta có: 

\(f'(x) = - \frac{{3a}}{{{{(x + 1)}^2}}} + b{e^x}(x + 1);\)

\(f'(0) = - 22 \Leftrightarrow - 3a + 2b = - 22\,(1)\)   

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\int\limits_0^1 {f(x)dx = 5} \\
 \Leftrightarrow \int\limits_0^1 {\left( {a{{\left( {x + 1} \right)}^{ - 3}} + bx{e^x}} \right)dx}  = 5
\end{array}\\
{ \Leftrightarrow \left. {\frac{a}{{ - 2{{(x + 1)}^2}}}} \right|_0^1 + b\left( {\left. {x{e^x}} \right|_0^1 - \int\limits_0^1 {{e^x}dx} } \right) = 5}\\
\begin{array}{l}
 \Leftrightarrow \left. {\frac{{ - a}}{{2{{(x + 1)}^2}}}} \right|_0^1 + \left. {bx{e^x}} \right|_0^1 - \left. {b{e^x}} \right|_0^1\\
 \Leftrightarrow \frac{3}{8}a + b = 5{\mkern 1mu} {\mkern 1mu} (2)
\end{array}
\end{array}\)

Từ (1), (2) suy ra \(a=8; b=2\).