Massey logo
Home > College of Sciences > Institute of Fundamental Sciences >
Maths First > Online Maths Help > Calculus > Integration > Integration by Substitution 2
SEARCH
MASSEY
MathsFirst logo College of Science Brandstrip
  Home  |  Study  |  Research  |  Extramural  |  Campuses  |  Colleges  |  About Massey  |  Library  |  Fees  |  Enrolment

 

Integration

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 the indefinite integral of if we replaceg of xby uandg dash of x dxbydu.

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:

the indefinite integral of f(ax2 + b)xm dx

the indefinite integral of f(ax2 + b)xm dx where m is odd can be rewriiten as the indefinite integral of f(ax2 + b)xm − 1 x dx

and the substitution u = ax2 + b used.

Example

Let us consider the indefinite integral ofx3(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 the indefinite integral of½x2(x2 + 1) 99 2x dx.

Since u = x2 + 1, x2 = u − 1

Hence the integral becomes the indefinite integral of½(u − 1)u99 du = ½the indefinite integral of(u100u99) du

This is integrable.

½the indefinite integral of(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:

indefinite integral of 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 indefinite integral of x times the square root of (2x plus 1).

The square root is a composite function with inner function 2x + 1. Let u = sqrt(2x+1) 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 one half the integral of x times the square root of (2x  plus 1) times 2  with respect to x

Hence the integral becomes one half the indefinite integral of (u minus 1) over 2 times the square root of u du

Integrating and substituting for u: a quarter (u to the power (5/2) divided by (5/2) minus u to the power (3/2) divided by (3/2)) plus ca quarter ((2x plus 1) to the power (5/2) divided by (5/2) minus (2x plus 1)  to the power (3/2) divided by (3/2)) plus c

 

Exercises

indefinite integral of dx =

<< Integration by Substitution 1 | Integration Index | Integration by Trigonometric Substitution 1 >>

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