$\int x^n e^{cx}\; \mathrm{d}x = \frac{1}{c} x^n e^{cx} - \frac{n}{c}\int x^{n-1} e^{cx} \mathrm{d}x = \left( \frac{\partial}{\partial c} \right)^n \frac{e^{cx}}{c} = e^{cx}\sum_{i=0}^n (-1)^i\,\frac{n!}{(n-i)!\,c^{i+1}}\,x^{n-i} = e^{cx}\sum_{i=0}^n (-1)^{n-i}\,\frac{n!}{i!\,c^{n-i+1}}\,x^i$
In the above equation, can someone explain to me what happens after the integration by parts?
On
What he (?) did, is a simple application of the differentiation under the integral sign "a posteriori", namely he found a recursive way to calculate that integral, which is by series, and he said like "hey, integrating by parts $n$ times is boring, just observe that differentiating $n$ times with respect to $c$ (which becomes a variable of differentiation instead of a constant) you will get the starting integral".
Indeed you can reach that writings by minding again "a posteriori". Indeed:
$$\frac{\text{d}}{\text{d}c}\frac{e^{cx}}{c} = \frac{xe^{cx}}{c} - \frac{e^{cx}}{c^2}$$
Again
$$\frac{\text{d}^2}{\text{d}c^2}\frac{e^{cx}}{c} = \frac{x e^{c x}}{c}-\frac{e^{c x}}{c^2}$$
Again
$$\frac{\text{d}^3}{\text{d}c^3}\frac{e^{cx}}{c} = \frac{2 e^{c x}}{c^3}-\frac{2 x e^{c x}}{c^2}+\frac{x^2 e^{c x}}{c}$$
And again
$$\frac{\text{d}^4}{\text{d}c^4}\frac{e^{cx}}{c} = -\frac{6 e^{c x}}{c^4}+\frac{6 x e^{c x}}{c^3}-\frac{3 x^2 e^{c x}}{c^2}+\frac{x^3 e^{c x}}{c}$$
The more derivatives you compute, the more you will see a pattern which will end up to be exactly
$$\sum_{i = 0}^n (-1)^i \frac{1}{c^{i+1}}\left(\frac{n!}{(n-i)!}\right)x^{n-i}$$
Then you can rename the variable by shifting and obtaining the final result.
More Derivative
$$\frac{\text{d}^5}{\text{d}c^5}\frac{e^{cx}}{c} = \frac{24 e^{c x}}{c^5}-\frac{24 x e^{c x}}{c^4}+\frac{12 x^2 e^{c x}}{c^3}-\frac{4 x^3 e^{c x}}{c^2}+\frac{x^4 e^{c x}}{c}$$
$$\frac{\text{d}^6}{\text{d}c^6}\frac{e^{cx}}{c} = -\frac{120 e^{c x}}{c^6}+\frac{120 x e^{c x}}{c^5}-\frac{60 x^2 e^{c x}}{c^4}+\frac{20 x^3 e^{c x}}{c^3}-\frac{5 x^4 e^{c x}}{c^2}+\frac{x^5 e^{c x}}{c}$$
$$\frac{\text{d}^7}{\text{d}c^7}\frac{e^{cx}}{c} = \frac{720 e^{c x}}{c^7}-\frac{720 x e^{c x}}{c^6}+\frac{360 x^2 e^{c x}}{c^5}-\frac{120 x^3 e^{c x}}{c^4}+\frac{30 x^4 e^{c x}}{c^3}-\frac{6 x^5 e^{c x}}{c^2}+\frac{x^6 e^{c x}}{c}$$
$$\frac{\text{d}^8}{\text{d}c^8}\frac{e^{cx}}{c} = -\frac{5040 e^{c x}}{c^8}+\frac{5040 x e^{c x}}{c^7}-\frac{2520 x^2 e^{c x}}{c^6}+\frac{840 x^3 e^{c x}}{c^5}-\frac{210 x^4 e^{c x}}{c^4}+\frac{42 x^5 e^{c x}}{c^3}-\frac{7 x^6 e^{c x}}{c^2}+\frac{x^7 e^{c x}}{c}$$
$$\frac{\text{d}^9}{\text{d}c^9}\frac{e^{cx}}{c} = \frac{40320 e^{c x}}{c^9}-\frac{40320 x e^{c x}}{c^8}+\frac{20160 x^2 e^{c x}}{c^7}-\frac{6720 x^3 e^{c x}}{c^6}+\frac{1680 x^4 e^{c x}}{c^5}-\frac{336 x^5 e^{c x}}{c^4}+\frac{56 x^6 e^{c x}}{c^3}-\frac{8 x^7 e^{c x}}{c^2}+\frac{x^8 e^{c x}}{c}$$
$$\frac{\text{d}^{10}}{\text{d}c^{10}}\frac{e^{cx}}{c} = -\frac{362880 e^{c x}}{c^{10}}+\frac{362880 x e^{c x}}{c^9}-\frac{181440 x^2 e^{c x}}{c^8}+\frac{60480 x^3 e^{c x}}{c^7}-\frac{15120 x^4 e^{c x}}{c^6}+\frac{3024 x^5 e^{c x}}{c^5}-\frac{504 x^6 e^{c x}}{c^4}+\frac{72 x^7 e^{c x}}{c^3}-\frac{9 x^8 e^{c x}}{c^2}+\frac{x^9 e^{c x}}{c}$$
$$\frac{\text{d}^{11}}{\text{d}c^{11}}\frac{e^{cx}}{c} = \frac{3628800 e^{c x}}{c^{11}}-\frac{3628800 x e^{c x}}{c^{10}}+\frac{1814400 x^2 e^{c x}}{c^9}-\frac{604800 x^3 e^{c x}}{c^8}+\frac{151200 x^4 e^{c x}}{c^7}-\frac{30240 x^5 e^{c x}}{c^6}+\frac{5040 x^6 e^{c x}}{c^5}-\frac{720 x^7 e^{c x}}{c^4}+\frac{90 x^8 e^{c x}}{c^3}-\frac{10 x^9 e^{c x}}{c^2}+\frac{x^{10} e^{c x}}{c}$$
Differentiation under the integral sign gives $$ \int x^ne^{cx}dx =\int \frac{∂^n}{∂c^n}e^{cx}dx =\frac{∂^n}{∂c^n}\int e^{cx}dx =\frac{∂^n}{∂c^n}\frac{e^{cx}+D(c)}c $$ where $D(c)$ is the integration constant that may depend on $c$.