Solution to Exercise 11.6-2


© 2001 by Karl Hahn 		KCT logo

The problem is to find the indefinite integral of 


   


   x2 dx
  ________
 √7x2 - 21

Step 1: Use the trick to do polynomial long division on this integrand. Then use the results on equation 11.5-32 to break this integral up into two pieces. When you've done that, click here.

The trick would convert this integral into 


   


   x2 dx
  ________  =
 √7x2 - 21


  


    x2
         
 7x2 - 21


  ________
 √7x2 - 21 dx

Your polynomial long division would be 

              1
               
              7              
   7x2 - 21 ) x2  +  0x  +  0
              x2         -  3
                             
                            3
which gives a quotient of  1/7  and a remainder of 3. Applying those results and equation 11.5-32 to the integrand and then simplifying gives 


   


    x2
         
 7x2 - 21


  ________
 √7x2 - 21 dx  =



  1
   
  7


 


  ________
 √7x2 - 21 dx  +  3



 


     dx
  ________
 √7x2 - 21

Step 2: Factor the constant outside the radical. But which constant? Remember that you want the constant term under the radical to be a 1. Observe what the radical does to the constant as it emerges. When you've got this step done, click here.

You have to factor 21 from what's under each radical to get it into standard form. Bringing the 21 outside the radical makes us take its square root. Since everything is being done in sevens and threes in this problem, I have factored 21 into  7 × 3,  but you don't have to do that. It just makes things easier to keep track of in what follows. 


   1
    
   7


 


  ________
 √7x2 - 21 dx  +  3



 


     dx
  ________  =
 √7x2 - 21


           _  _
          √3 √7
               
            7




 









          3
 dx  +   _  _
        √7 √3



 


     dx

Observe that the multipliers out ahead of each of the integrals are actually equal to each other and indeed are both equal to 

   _
  √3
   _
  √7

Step 3: Look up the form on the table and make the indicated substitutions. When you've done that, click here.

Your substitutions are 


                x
   cosh(u)  =   _
               √3
and 
 _
√3 sinh(u) du  =  dx

and the integral becomes 


    3
    _
   √7


 


  ____________
 √cosh2(u) - 1 sinh(u) du  +



   3
   _
  √7


 


   sinh(u) du
  ____________
 √cosh2(u) - 1

Step 4: Use hyperbolic identities to put these integrands into a form you know how to integrate. Use equations 11.6-8b, and 11.6-13a to put this stuff into shape, then click here.

From equation 11.6-8b and doing the simplifications an cancellations that follow (and again dropping the ±), you get 


    3
    _
   √7


 


  ____________
 √cosh2(u) - 1 sinh(u) du  +



   3
   _
  √7


 


   sinh(u) du
  ____________  =
 √cosh2(u) - 1

                                         3
                                         _
                                        √7

 


 sinh2(u) du  +


   3
   _
  √7

 


 du

and applying equation 11.6-13a to that you get 


    3
    _
   √7


 



 sinh2(u) du  +



   3
   _
  √7


 



 du  =



   3
    _
  2√7


 



 (cosh(2u) - 1) du  +



   3
   _
  √7


 



 du

You can simplify that even further by observing that there are like terms in the two integrals. Gathering them into one integral gives 


    3
     _
   2√7


 



 (cosh(2u) - 1) du  +



   3
   _
  √7


 



 du  =



   3
    _
  2√7


 



 cosh(2u) du  +



   3
    _
  2√7


 



 du

Step 5: Integrate. Remember that hyperbolic arcsine and hyperbolic arccosine are antiderivatives of each other. When you've integrated this, click here.

You should have 


    3
     _
   2√7


 



 cosh(2u) du  +



   3
    _
  2√7


 



 du  =



   3
    _ sinh(2u)  +
  4√7


   3
    _ u  +  C
  2√7

Step 6: Substitute back. Your substitution was 

                x
   cosh(u)  =   _
               √3
Note that you will have to use the hyperbolic arccosine function to substitute back the bare u. You'll also have to use equations 11.6-12a and 11.6-8b to whip that  sinh(2u)  term into a shape where you can substitute it back. Do all that, then click here, and you'll be done.

Applying equation 11.6-12a gives 


    3
     _ sinh(2u)  +
   4√7


   3
    _ u  +  C  =
  2√7


   3
    _ sinh(u)cosh(u)  +
  2√7


   3
    _ u  +  C
  2√7

Then applying equation 11.6-8b gives 


    3
     _ sinh(u)cosh(u)  +
   2√7


   3
    _ u  +  C  =
  2√7


   3
    _
  2√7


  ____________
 √cosh2(u) - 1 cosh(u)  +



   3
    _ u  +  C
  2√7

Substituting back to x you have 

    _
   √3x
     _
   2√7


 


       3
  +     _ arccosh
      2√7





 x
 _
√3






  +  C  =



                                     x
                                      
                                    14


  ________
 √7x2 - 21



      3
  +    _ arccosh
     2√7





 x
 _
√3






  +  C

That last simplification on the bottom line of the above equation isn't really necessary, but it does make the answer prettier. And you could take it one more step by writing the hyperbolic arccosine term as a log using equation 11.6b-4, but that is really unnecessary.

Of course this solution is no piece of cake to take the derivative of, but do it anyway for practice.

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