Antiderivative of $x^ae^\left( bx \right)$ I need to know the primitive function (Antiderivative) of this function:
$$f(x)=x^ae^\left( bx \right)$$
where $a$ and $b$ is a positive constants
please how could I find the primitive of  function ?  is there any technique concerning this types?
Thanks in advance
 A: The fundamental theorem of calculus states
$$\frac{d}{dx}\left(\int_{g(x)}^{\infty} h(t)\,dt\right)=-h(g(x))g'(x).\tag{1}$$
The definition of the Incomplete gamma function is
$$\Gamma(a,x) = \int_x^{\infty} t^{a-1}\,e^{-t}\, dt.\tag{2}$$
Combining these two gives
$$\frac{d}{dx}\Gamma(a+1,-bx)\stackrel{(2)}=\frac{d}{dx}\left(\int_{-bx}^{\infty} t^{a}\,e^{-t}\, dt\right)\stackrel{(1)}=-(-bx)^ae^{bx}(-b)=b(-b)^{a}x^ae^{bx}.\tag{3}$$
Therefore
\begin{align}
\int x^a e^{bx}\,dx&=\int b(-b)^{a}x^a e^{bx}b^{-1}(-b)^{-a}\,dx\\&\stackrel{(3)}=
b^{-1}(-b)^{-a}\int\left(\frac{d}{dx}\Gamma(a+1,-bx)\right)\,dx\\&=
b^{-1}(-b)^{-a}\Gamma(a+1,-bx)+C.
\end{align}
A: This is a standard integration question which can be solved by using reduction formulas (just a fancy application of making repeated integration by parts). Main idea is you use integration by parts, which will ultimately required differentiating $x^a$ reducing the index by 1, then you generalize the rule, call it $I_0$, repeatedly apply until you get to $I_0$. Other similar integrals which make use of this are: $e^{x} \sin(x)$
A: Note that if $a\in\mathbb{N}$, then
$$f(x)=x^ae^\left( bx \right)$$
can be written as
$$f(x)=\frac{\partial^a}{\partial b^a}e^{bx}.$$
So you can first find an an antiderivative of $e^{bx}$ and then differentiate it $a$ times w.r.t $b$.
Otherwise you need the incomplete gamma function as said by others.
