L&T, 19.2
This is the integral equivalent to the differential of a product. Start with
Integrate both sides with respect to ,
Now use and . Rearrange the terms, and find
This last equation is mainly used in the form
Example 5.10:
Evaluate .
Solution:
Put and . part: , therefore and . part: therefore and (constant of integration not needed here). Thus
Note that the part of the original integrand (i.e., ) was differentiated, but the part (i.e., ) was integrated. We obtained a new integral which was easier than the old one because was simpler than but was no harder than . It is a requirement that the resulting integral is no more complicated than the original!
Example 5.11:
Evaluate .
Solution:
Put , . , therefore . , therefore . We thus obtain
Now repeat this procedure: Put , . We find , and therefore . Finally and thus .
(We have put the constant of integration in at the end.)
Example 5.12:
Evaluate
.
Solution:
Even though this does not look like integration by parts, we can use a trick! Use the fact that the derivative of the logarithm is much more manageable than the logarithm itself, and use a function with derivative . Thus , , , , , .
Example 5.13:
Find .
Solution:
Here we can integrate or differentiate , and differentiate or integrate , since the integrals and derivatives of both functions are as simple as the original function. We choose , therefore and .
Now integrate by parts again.
Note: Initially we differentiated , taking as a derivative. We must use the same procedure again, and not switch and . I.e., we must put and . Therefore , , and thus , . It follows that , and so .
Bring to the left-hand side,
and thus, finally,