Question: Undoubtedly, this question is so easy but I'd like to ask it. We know that the way in which the indefinite integrals like $\int P(x)e^{ax}dx$ and $\int P(x)\sin(bx)dx$ wherein $P(x)$ is an arbitrary polynomial of $x$ and $a, b\in \mathbb R$ are evaluated is **Integrating by Parts** $\int udv=uv-\int vdu$.

So we should take $u$ and $dv$ frequently until the whole integral becomes an easy integral. Is there any fast method in which we can do above integrals without using the classic formula? Thank you for the help.

Answer: 
We know $$\int P(x)e^{ax}\,dx=Q(x)e^{ax}+C$$ for some polynomial $Q(x)$ of the same degree as $P(x)$ (the coefficients of $Q$ may also involve $a$). Differentiating, $P(x)e^{ax}=(aQ(x)+Q'(x))e^{ax}$, so $$P(x)=aQ(x)+Q'(x)$$ Now you can use this to find the coefficients of $Q$ one at a time, starting with the leading coefficient and working your way down the degrees. 

For example, to find $\int(3x^2+4x+5)e^{6x}\,dx$, let $Q(x)=rx^2+sx+t$; then $$aQ(x)+Q'(x)=6rx^2+(6s+2r)x+(6t+s)$$, so comparing coefficients we get $$3=6r;\qquad4=6s+2r;\qquad5=6t+s$$ and you get $r$ from the first equation, $s$ from the second, and $t$ from the third. 

For the trig integral, you can use Jack's idea (but I think Jack meant $e^{ibx}$), or you can say that the answer will be of the form $Q(x)\sin(bx)+R(x)\cos(bx)$ and solve as above, noting that of the two polynomials $Q$ and $R$ one has only even, the other only odd, degree terms.
With $Q$ and $P$ as in Gerry's answer, you can write $Q = a^{-1} P - a^{-2} P' + a^{-3} P'' - \ldots$ (stopping when the derivatives become $0$). Thus in his example with 
$P = 3 x^2 + 4 x + 5$ and $a = 6$, 
$$ Q = \frac{3 x^2 + 4 x + 5}{6} - \frac{6 x + 4}{6^2} + \frac{6}{6^3} = \frac{x^2}{2} + \frac{x}{2} + \frac{3}{4}$$

Similarly, $\int P(x) \sin(bx)\ dx = Q(x) \sin(bx) + R(x) \cos(bx)$ where 
$$ \eqalign{Q &= b^{-2} P' - b^{-4} P''' + b^{-6} P^{(5)} - \ldots\cr
 R &= - b^{-1} P + b^{-3} P'' - b^{-5} P^{(4)} + \ldots\cr}$$