Answer: \(\displaystyle \mathrm f(x)=\frac23e^{3x}+\frac23x^3+3x+1\).

The key point is to differentiate first, then substitute the required value or solve the resulting equation.

Given

integrate once:

The gradient is \(5\) at \(x=0\), so

\(5=2e^0+0+C.\)

Thus \(C=3\), and

Integrate again:

\(\displaystyle y=\frac23e^{3x}+\frac23x^3+3x+D.\)

The curve passes through \(\left(0,\frac53\right)\), so

\(\displaystyle \frac53=\frac23+D.\)

Therefore \(D=1\), and hence