On-line Math 21

5.4  Techniques of integration

Example 1

ó õ x Ö   x2+4   dx.

Solution

I would take u(x) to be u(x) = x²+4 . I'm looking for an ``inside part'' (remember the chain rule), and I hope I can find the essential part of its derivative to use to make du . But, actually, that last part is fairly automatic. Note that

du dx = 2x,

so du = 2xdx , or du/2x = dx :

ó õ x Ö   x2+4   dx = ó õ xÖudu/2x = 1 2 ó õ Öudu = 1 2 u3/2/(3/2)+C = 1 3 (x2+4)3/2+C

That x outside the square root was essential for this method to work, since I needed to be able to replace the dx in terms of u and du , but du , in terms of x , has that extra x .

At first, I recommend you do all substitution integrals this way: Find what you want u to be (looking for the du to be there (except for constants)), then find du in terms of x and dx . Then substitute the u for the inside stuff, and replace the dx in terms of du and x . If your substitution was going to work, the x dependencies will all vanish. Then, integrate the resulting formula in terms of u as a variable, and finally substitute back. As you get more comfortable with the method, though, you will easily group together the terms that need to be there to be du , up to the constants, which are easy to deal with. There is one caveat with that, however, in that it is easy to ``correct for'' the constants incorrectly. Solving for dx as above automatically takes care of the constants.

Copyright (c) 2000 by David L. Johnson.