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 replace by and by .
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.
Example
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
Exercise
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.
Example
Evaluate .
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:  
Exercises
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