Recurrent IPPs and recurrence relations

When a single application of the IPP is not enough

Certain integrals, such as the integral of x² e^x dx or, more generally, the integral of x^n e^x dx, require the IPP to be applied several times in succession, as each application only reduces the degree of the polynomial by 1.

Step-by-step example

Calculate I2 = the integral of x² e^x dx. Step 1: u = x², dv = e^x dx → du = 2x dx, v = e^x. I2 = x^2 e^x – integral of 2x e^x dx = x^2 e^x – 2 * integral of x e^x dx. Step 2: we use the known result that the integral of x e^x dx = (x–1) e^x + C. I2 = x^2 e^x – 2(x–1) e^x + C = (x^2 – 2x + 2) e^x + C.

General recurrence formula

For In = integral of x^n e^x dx, the IPP (u = x^n, dv = e^x dx) gives: In = x^n e^x - n * I(n-1)

This relationship allows us to calculate $In$ by successive approximation starting from $I_0 = e^x + C$, without having to repeat the entire calculation each time.

Table of the first terms

n	$In$ (to within a constant)
0	e^x
1	(x-1) e^x
2	(x^2-2x+2) e^x
3	(x^3-3x^2+6x-6) e^x

Common pitfall

At each step, you must maintain the same role for u (the polynomial factor being differentiated); otherwise, the degree does not decrease and the calculation loops indefinitely instead of terminating.