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 \[\begin{aligned} I_0&= \frac{1}{3}e^{3x}+c_0\\[0.25cm] I_1&=\frac{1}{3}xe^{3x}-\frac{1}{3}I_0=\frac{1}{9}(3x-1)e^{3x}+c_1\\[0.25cm]
I_2&=\frac{1}{3}x^2e^{3x}-\frac{2}{3}I_1=\frac{1}{27}(9x^2-6x+2)e^{3x}+c_2\\[0.25cm]
I_3&=\frac{1}{3}x^3e^{3x}-I_2=\frac{1}{27}(9x^3-9x^2+6x-2)e^{3x}+c_3\\[0.25cm]
I_4&=\frac{1}{3}x^3e^{3x}-\frac{4}{3}I_3=\frac{1}{81}(27x^4-36x^3+36x^2-24x+8)e^{3x}+c_4\end{aligned}\]