Worked Example 11.8-2 © 2001 by Karl Hahn

2) The example is to integrate

dx (x2 + 9)2

Before making the trig substitution, we have to divide out the constant

dx = (x2 + 9)2

1 81

dx eq. 11.8-34 ((x/3)2 + 1)2

              x
   tan(u)  =   
              3

3 (tan2(u) + 1) du  =  dx                 eq. 11.8-35a

    1
     
   27

   tan2(u) + 1
                du  =
 (tan2(u) + 1)2

   1
    
  27

      du
                                eq. 11.8-35b
 tan2(u) + 1

                 1
   sec(u)  =        
              cos(u)

    1
     
   27

      du
              =
 tan2(u) + 1

   1
    
  27

    du
          =
 sec2(u)

   1
    
  27

 cos2(u) du         eq. 11.8-35c

    1
     
   27

 cos2(u) du  =

   1
    
  54

 (1 + cos(2u)) du  =                  eq. 11.8-35d

          1
            u  +
         54

   1
      sin(2u)  +  C  =
  108

   1        1
     u  +     sin(u)cos(u)  +  C
  54       54

                 1
   cos(u)  =          =
              sec(u)

        1
   ___________                             eq. 11.8-36a
  √tan2(u) + 1

              x
   tan(u)  =   
              3

     dx
            =
 (x2 + 9)2

  1
    arctan
 54

x
 
3

      1     x/3
  +                 +  C  =        eq. 11.8-36b
     54 (x/3)2 + 1

                                  1
                                    arctan
                                 54

x
 
3

      1    x
  +             +  C
     18 x2 + 9

An Alternative Method

You can use a cute variation on integration by parts to integrate anything in the form of

dx (x2 + a2)n

whenever n is an integer greater than 1. First, using simple algebra, convert it to:

1 a2n

dx ((x/a)2 + 1)n

and then, putting in  s = x/a  and  a ds = dx,  it becomes:

1 a2n-1

ds (s2 + 1)n

So much for the easy part. The trick now is to integrate the new integrand using integration by parts. The parts to choose are

1 u = (s2 + 1)n

and

dv = ds

By differentiating u and integrating dv, we find that:

-2n s ds du = (s2 + 1)n+1

and

v = s

We apply the integration-by-parts formula:

u dv = uv -

v du

When you do that with the parts we established above, you get

ds = (s2 + 1)n

s + 2n (s2 + 1)n

s2 ds (s2 + 1)n+1

On the surface, this does not seem to have gotten any better. Indeed it now seems worse than the original because the power in the denominator on the right has increased by one, and we now also have that nasty s² on top.

The first trick is to get rid of the s² in the numerator of the integral on the right. We use a little algebra for that. If you both add and subtract 1 from that numerator, isn't it true that you haven't changed it a bit?? That's what we do:

ds = (s2 + 1)n

s + 2n (s2 + 1)n

(s2 + 1 - 1) ds = (s2 + 1)n+1

s + 2n (s2 + 1)n

(s2 + 1) ds - (s2 + 1)n+1

ds (s2 + 1)n+1

Now take the cancellation offered in the second-to-last integral and distribute the 2n to get:

ds = (s2 + 1)n

s + 2n (s2 + 1)n

ds - 2n (s2 + 1)n

ds (s2 + 1)n+1

So we're getting closer. Notice that the integral to the left of the equal is the same as the second-to-last integral. That means this is a little like the gotcha we had to contend with when we integrated inverse functions and when we integrated  e^xsin(x).  But this is even worse. You could go ahead and solve for the integral on the left, but then you'd have the integral with the n power in terms of a similar integral with an n+1 power. That isn't very helpful because as you iterate this to find the right-hand integral, the powers just keep getting larger and larger.

But what if you solve for the integral that has the n+1 power? Then you have the integral with the n+1 power in terms of a similar integral with an n power. Now you have powers that are going down instead of up. So what we do here is take the solution we used on the gotcha problems, and stand it on its head. Gathering terms and moving the n+1 power integral over to the left, you get:

2n

ds = (s2 + 1)n+1

s + (2n-1) (s2 + 1)n

ds (s2 + 1)n

And solving for the integral on the left, you have:

ds = (s2 + 1)n+1

1 2n

s + (s2 + 1)n

2n-1 2n

ds (s2 + 1)n

If this is true when you have n+1 as the power on the left, then it must also be true when you replace n+1 with n and replace n with n-1 (do you see why?):

ds = (s2 + 1)n

1 2n-2

s + (s2 + 1)n-1

2n-3 2n-2

ds (s2 + 1)n-1

which will work whenever n is greater than 1. But when  n = 1,  we already know that the integral is  arctan(s) + C.  So whenever you have an integral like this, you just iterate this formula until you get down to an exponent of 1, and then you know how to integrate what remains.

I'll leave it to you to substitute back to the integral in terms of x (don't forget the  1/a^2n-1  multiplier that we had at the beginning).

You can use the same method to integrate

x2 dx (x2 + 1)n

Just observe that (according to the algebra trick we've already used once here):

x2 = (x2 + 1)n

1 - (x2 + 1)n-1

1 (x2 + 1)n

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