On-line Math 21

5.4  Techniques of integration

Example 2

ó õ x2 x3+7 dx

Solution

Here, the denominator is the ``inside''. The first example as a bit clearer on what was the inside term, the prime candidate for u , but as you'll see denominators are also likely suspects. So, set

u = x3+7,

and then

du = 3x2dx, or 1 3x2 du = dx.

Supstitute in terms of u and du :

ó õ x2 x3+7 dx = ó õ x2 u 1 3x2 du = 1 3 ó õ du u = 1 3 ln|u|+C.

At this stage I always ask myself: ``Now, what was u ?'' u = x³+7 . so

ó õ x2 x3+7 dx = 1 3 ln|u|+C = 1 3 ln| x3+7| +C.

Copyright (c) 2000 by David L. Johnson.