Example 1 Suppose you want to calculate the integral $\int x^4e^{3x}\,\dd x$. Then you can repeat integration by parts (four times) to get to the final result, because the exponent of the power of $x$ in the integrand is then in each step reduced by $1$. But this repeated calculation is elaborate and time consuming. More convenient is to generalise the problem and derive a reduction formula.

For $n\ge 0$ we define the integral $I_n$ as $$I_n=\int x^ne^{ax}\,\dd x$$ Note that $$I_0 = \int e^{ax}\,\dd x=\frac{1}{a}e^{ax}+c_0$$ for some constant $c_0$.

By integration by parts, we find a formula for $I_n$ in terms of $I_{n-1}$ :

If $u=x^n$ and $dv=e^{ax}\,\dd x$, then $\dd u=n\,x^{n-1}\,\dd x$ and $v=\int \dd v= \frac{1}{a}e^{ax}$, and therefore $$\begin{aligned}I_n&= \int x^n\,e^{ax}\,\dd x \\[0.25cm] &\phantom{abcdwxyz} \blue{\left[\int u\,\dd v= u\,v -\int v\,\dd u\right]} \\ &\phantom{abcdwxyz}\blue{\text{rule for integration by parts}}\\[0.25cm] &= x^n\cdot \frac{1}{a}e^{ax}-\int \frac{1}{a}e^{ax}\cdot n\,x^{n-1}\,\dd x\\ &\blue{\phantom{abcdwxyz}\text{application of the rule of integration by parts}}\\[0.25cm] &= \frac{1}{a}x^ne^{ax}-\frac{n}{a}\int x^{n-1}e^{ax}\,\dd x\\ &\blue{\phantom{abcdwxyz}\text{rewriting}}\\[0.25cm] &=\frac{1}{a}x^ne^{ax}-\frac{n}{a}I_{n-1}\\ &\blue{\phantom{abcdwxyz}\text{reduction formula}}\end{aligned}$$
We can now apply the reduction formula $$I_n=\frac{1}{a}x^ne^{ax}-\frac{n}{a}I_{n-1}$$ to $n=4$ and $a=3$ and get $$\begin{aligned} I_0&= \frac{1}{3}e^{3x}+c_0\\[0.25cm] I_1&=\frac{1}{3}xe^{3x}-\frac{1}{3}I_0=\frac{1}{9}(3x-1)e^{3x}+c_1\\[0.25cm] I_2&=\frac{1}{3}x^2e^{3x}-\frac{2}{3}I_1=\frac{1}{27}(9x^2-6x+2)e^{3x}+c_2\\[0.25cm] I_3&=\frac{1}{3}x^3e^{3x}-I_2=\frac{1}{27}(9x^3-9x^2+6x-2)e^{3x}+c_3\\[0.25cm] I_4&=\frac{1}{3}x^3e^{3x}-\frac{4}{3}I_3=\frac{1}{81}(27x^4-36x^3+36x^2-24x+8)e^{3x}+c_4\end{aligned}$$