Solution to Exercise 11.6-1


© 2001 by Karl Hahn 		KCT logo

The problem is to use completing the square and hyperbolic substitution to find the indefinite integral: 


   


      x dx
  _____________
 √x2 - 10x + 41

Step 1: Complete the square. Half the middle coefficient is -5. So when you add that to x, what do you get for the substitution? Now square that expression and what do you have to add to it to get back the original quadratic,  q(x) = x2 - 10x + 41,  that's under the radical? When you've figured all that out, click here.

Completing the square gives you 


   v  =  x - 5
and 

v2  =  x2 - 10x + 25

Clearly you have to add 16 to that to get back to  q(x),  that is,  v2 + 16 = x2 - 10x + 41 = q(x).  And since  v = x - 5,  then  dv = dx,  and  x = v + 5. 

Step 2: Make the completed square substitution. Everything you need is given above. When you have done this step, click here.

You should have 


   


      x dx
  _____________  =
 √x2 - 10x + 41


  


 (v + 5) dv
   _______ 
  √v2 + 16

Step 3: Break it up into a sum of integrals. This is pretty easy (use equation 11.2-13b). Do it and then click here.

You should have 


   


 (v + 5) dv
   _______   =
  √v2 + 16


  


   v dv
  _______  +  5
 √v2 + 16


 


    dv
  _______
 √v2 + 16

Step 4: Determine the status of each of the summands. The first of the two summands will go with simple substitution. Come up with substitutions for that one that turn the integrand in v and dv into and expression in s and ds.

The second summand won't go with simple substitution. So factor the constant out of it. Do that and the simple substitution, then click here.

Your simple substitution would be 



   s  =  v2 + 16
and 


ds  =  2v dv
or 

1
  ds  =  v dv
2

You should have 


   


   v dv
  _______  +  5
 √v2 + 16


 


    dv
  _______  =
 √v2 + 16


  1
   
  2


 


 ds
  _  +
 √s


  5
   
  4


 


      dv
  __________
 √(v/4)2 + 1

Step 5: For the integral on the right, go to the table and see what form matches what you've got here. Prepare the corresponding substitution and rewrite the integral with it. When you've done all that click here.

Your substitution is 


               v
   sinh(u)  =   
               4
and 


4 cosh(u) du  =  dv
Your new integral is 


   1
    
   2


 


 ds
  _  +
 √s


  5
   
  4


 


      dv
  __________  =
 √(v/4)2 + 1


  1
   
  2


 


 ds
  _  +
 √s



  5



 


   cosh(u) du
  ____________
 √sinh2(u) + 1

Step 6: Use equation 11.6-8a to simplify. Then click here.

Now you should have 


   1
    
   2


 


 ds
  _  +
 √s



  5



 


   cosh(u) du
  ____________  =
 √sinh2(u) + 1


  1
   
  2


 


 ds
  _  +
 √s



  5



 

  ____________
 √sinh2(u) + 1
  ____________ du  =
 √sinh2(u) + 1


                                                        1
                                                         
                                                        2


 


 ds
  _  +
 √s



  5



 



 du

Step 7: Integrate. The version on the second line above is all integrals you know how to do. So do them. Then click here.

You should have 


   1
    
   2


 


 ds
  _  +
 √s



  5



 


         _
 du  =  √s  +  5u  +  C

Step 8: Substitute back to v. Recall that  s = v2 + 16  and  sinh(u) = v/4.  To use the latter back-substitution on the bare u you will have to employ an inverse hyperbolic function. Work all of that out and then click here.

You should have 


    _
   √s  +  5u  +  C  =



   _______
  √v2 + 16  +  5 arcsinh






v
 
4






  +  C

Step 9: Substitute back to x. This step is so easy that I'm not going to make you click for it. From your original substitution you have  v = x - 5,  so everywhere you see a v, put in a  x - 5.  You end up with 


    _______
   √v2 + 16  +  5 arcsinh






v
 
4






  +  C  =



   _____________
  √x2 - 10x + 41  +  5 arcsinh






x - 5
     
  4






  +  C

If you're feeling ambitious you can replace the  arcsinh  expression using equation 11.6b-6, but most instructors won't require that you do that. The solution you should check by taking its derivative, however, is the one shown above. Click here to see the derivative of the hyperbolic arcsine.

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