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## 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)xmdx 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 (u100u99) du

This is integrable.

½ (u100u99) du = ½((1/101)u101 −(1/100)u100) + c = (1/202)(x2 + 1)101 − (1/200)(x2 + 1)100 + c

### Exercise

Evaluate the indefinite integral: dx =

## 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  dx = Contact Us | About Massey University | Sitemap | Disclaimer | Last updated: November 21, 2012     © Massey University 2003