Integration by Substitution 2 - Harder Algebraic Substitution
We assume that you are familiar with the material in integration by substitution 1.
Recall the Substitution Rule
To integrate if we replaceby andby.
This converts the original integral into a simpler one. We extend this idea to some harder integrals.
Sometimes an integrand may need a bit of algebraic manipulation to make it integrable. Make sure that the integral with respect to the variable u does not contain any x.
Consider some special cases:
f(ax2 + b)xm dx
f(ax2 + b)xm dx where m is odd can be rewriiten as f(ax2 + b)xm − 1 x dx
and the substitution u = ax2 + b used.
Let us consider x3(x2 + 1)99dx.
(x2 + 1)99 is a composite function. One would consider replacing the inner function (x2 + 1) by u.
Thus du = 2x dx, which can be obtained from x3 since it equals x2 times x.
We rewrite the integral as ½x2(x2 + 1) 99 2x dx.
Since u = x2 + 1, x2 = u − 1
Hence the integral becomes ½(u − 1)u99 du = ½(u100 − u99) du
This is integrable.
½(u100 − u99) du = ½((1/101)u101 −(1/100)u100) + c = (1/202)(x2 + 1)101 − (1/200)(x2 + 1)100 + c
Integrand contains (mx + b)(m/n)
When the integrand contains an expression of the form (mx + b)(m/n) then the substitution u = (mx + b)(1/n) is often suitable. Thus un = mx + b and nun − 1 du = m dx.
The square root is a composite function with inner function 2x + 1. Let u = which is equivalent to u2 = 2x + 1
Then 2u du = 2 dx or u du = dx. We need to express the factor x in terms of u.
u2 = 2x + 1 so x = ½(u2 − 1)
We can write the integral as
Hence the integral becomes
Integrating and substituting for u: